Frank Solutions for Class 10 Maths Chapter 21 Trigonometric Identities

Students can refer to these Frank Solutions for Class 10 Maths Chapter 21 Trigonometric Identities, which help to gain knowledge. So, it is best to address these solutions for better guidance. Solving the exercises in each chapter will ensure that students score good marks in the exams. Our solution module utilizes various shortcut tips and practical examples to explain all the exercise questions in a simple and easily understandable language.

Chapter 21 – Trigonometric Identities, are equalities that involve trigonometric functions and are true for every value of the occurring variables, where both sides of the equality are defined. Expert tutors at BYJU’S have designed these solutions in a very lucid and clear manner that helps students solve problems in the most efficient possible ways. Download pdf of Frank Solutions for Class 10 Maths Chapter 21 in their respective links.

Download the PDF of Frank Solutions for Class 10 Maths Chapter 21 Trigonometric Identities

 

frank solutions for class 10 maths chapter 21 trigonometric identities 01
frank solutions for class 10 maths chapter 21 trigonometric identities 02
frank solutions for class 10 maths chapter 21 trigonometric identities 03
frank solutions for class 10 maths chapter 21 trigonometric identities 04
frank solutions for class 10 maths chapter 21 trigonometric identities 05
frank solutions for class 10 maths chapter 21 trigonometric identities 06
frank solutions for class 10 maths chapter 21 trigonometric identities 07
frank solutions for class 10 maths chapter 21 trigonometric identities 08
frank solutions for class 10 maths chapter 21 trigonometric identities 09
frank solutions for class 10 maths chapter 21 trigonometric identities 10
frank solutions for class 10 maths chapter 21 trigonometric identities 11
frank solutions for class 10 maths chapter 21 trigonometric identities 12
frank solutions for class 10 maths chapter 21 trigonometric identities 13
frank solutions for class 10 maths chapter 21 trigonometric identities 14
frank solutions for class 10 maths chapter 21 trigonometric identities 15
frank solutions for class 10 maths chapter 21 trigonometric identities 16
frank solutions for class 10 maths chapter 21 trigonometric identities 17
frank solutions for class 10 maths chapter 21 trigonometric identities 18
frank solutions for class 10 maths chapter 21 trigonometric identities 19
frank solutions for class 10 maths chapter 21 trigonometric identities 20
frank solutions for class 10 maths chapter 21 trigonometric identities 21
frank solutions for class 10 maths chapter 21 trigonometric identities 22
frank solutions for class 10 maths chapter 21 trigonometric identities 23
frank solutions for class 10 maths chapter 21 trigonometric identities 24
frank solutions for class 10 maths chapter 21 trigonometric identities 25
frank solutions for class 10 maths chapter 21 trigonometric identities 26
frank solutions for class 10 maths chapter 21 trigonometric identities 27
frank solutions for class 10 maths chapter 21 trigonometric identities 28
frank solutions for class 10 maths chapter 21 trigonometric identities 29
frank solutions for class 10 maths chapter 21 trigonometric identities 30

 

Access answers to Frank Solutions for Class 10 Maths Chapter 21 Trigonometric Identities

1. Prove the following identities:

(i) (1 – sin2 θ) sec2θ = 1

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 – sin2 θ) sec2θ

We know that, (1 – sin2 θ) = cos2 θ

= cos2θ sec2θ

Also we know that, sec2θ = 1/cos2θ

= cos2 θ × (1/cos2θ)

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(ii) (1 – cos2 θ) sec2θ = tan2θ

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 – cos2 θ) sec2θ

We know that, (1 – cos2 θ) = sin2 θ

= sin2θ sec2θ

Also we know that, sec2θ = 1/cos2θ

= sin2 θ × (1/cos2θ)

= sin2θ/cos2θ

= tan2θ

Then, Right Hand Side (RHS) = tan2θ

Therefore, LHS = RHS

(iii) tan A + cot A = sec A cosec A

Solution:-

From the question first we consider Left Hand Side (LHS),

= tan A + cot A

We know that, tan A = sin A/cos A, cot A = cos A/ sin A

Then,

= (sin A/cos A) + (cos A/sin A)

= (sin2 A + cos2 A)/(sin A cos A)

Also we know that, sin2 A + cos2 A = 1

= 1/(sin A cos A)

= (1/sin A)(1/cos A)

= cosec A sec A

Then, Right Hand Side (RHS) = sec A cosec A

Therefore, LHS = RHS

(iv) sin θ(1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosec θ

Solution:-

From the question first we consider Left Hand Side (LHS),

= sin θ(1 + tan θ) + cos θ (1 + cot θ)

We know that, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ

= sin θ(1 + (sin θ/cos θ)) + cos θ (1 + (cos θ/sin θ))

= sin θ((cos θ + sin θ)/cos θ)) + cos θ ((sin θ + cos θ)/sin θ))

= cos θ + sin θ ((sin θ/cos θ) + (cos θ/sin θ))

= cos θ + sin θ ((1/sin θ)(1/cos θ))

= sec θ + cosec θ

Then, Right Hand Side (RHS) = sec θ + cosec θ

Therefore, LHS = RHS

(v) (1 + cot θ – cosec θ) (1 + tan θ + sec θ) = 2

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + cot θ – cosec θ) (1 + tan θ + sec θ)

We know that,

cot θ = sin θ/cos θ, cosec θ = 1/cos θ, tan θ = cos θ/sin θ, sec θ = 1/sin θ

= (1 + (sin θ/cos θ) + (1/cos θ)) (1 + (cos θ/sin θ) – (1/sin θ))

Taking LCM we get,

= ((cos θ + sin θ + 1)/cos θ) ((sin θ + cos θ – 1)/sin θ)

= ((sin θ + cos θ)2 – 12)/(sin θ cos θ)

= (1 + 2 sin θ cos θ – 1)/(sin θ cos θ)

= (2 sin θ cos θ)/(sin θ cos θ)

By simplification we get,

= 2

Then, Right Hand Side (RHS) = 2

Therefore, LHS = RHS

(vi) sin θ cot θ + sin θ cosec θ = 1 + cos θ

Solution:-

From the question first we consider Left Hand Side (LHS),

= sin θ cot θ + sin θ cosec θ

We know that, cot θ = cos θ/sin θ, cosec θ = 1/sin θ

= sin θ (cos θ/sin θ) + sin θ (1/sin θ)

= cos θ + 1

Then, Right Hand Side (RHS) = 1 + cos θ

Therefore, LHS = RHS

(vii) sec A (1 – sin A) (sec A + tan A) = 1

Solution:-

From the question first we consider Left Hand Side (LHS),

= sec A (1 – sin A) (sec A + tan A)

We know that, sec A = 1/cos A, tan A = sin A/cos A

= (1/cos A) (1 – sin A) ((1/cos A) + (sin A/cos A))

= ((1 – sin A)/cos A) ((1 + sin A)/cos A)

By simplification we get,

= (1 – sin2A)/cos2 A

= cos2 A/cos2 A

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(viii) sec A (1 + sin A) (sec A – tan A) = 1

Solution:-

From the question first we consider Left Hand Side (LHS),

= sec A (1 + sin A) (sec A – tan A)

We know that, sec A = 1/cos A, tan A = sin A/cos A

= (1/cos A) (1 + sin A) ((1/cos A) – (sin A/cos A))

= ((1 + sin A)/cos A) ((1 – sin A)/cos A)

By simplification we get,

= (1 – sin2A)/cos2 A

= cos2 A/cos2 A

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(ix) cosec θ (1 + cos θ) (cosec θ – cot θ) = 1

Solution:-

From the question first we consider Left Hand Side (LHS),

= cosec θ (1 + cos θ) (cosec θ – cot θ)

We know that, cosec = 1/sin θ, cot θ = cos θ/sin θ

= (1/sin θ) (1 + cos θ) ((1/sin θ) – (cos θ/sin θ))

= ((1 + cos θ)/sin θ) ((1 – cos θ)/sin θ)

= ((1 – cos2 θ)/sin2 θ)

= sin2 θ/sin2 θ

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(x) (sec A – 1)/(sec A + 1) = (1 – cos A)/(1 + cos A)

Solution:-

From the question first we consider Left Hand Side (LHS),

= (sec A – 1)/(sec A + 1)

We know that, sec A = 1/cos A

= ((1/cos A) – 1)/((1/cos A) + 1)

By simplification we get,

= (1 – cos A)/(1 + cos A)

Then, Right Hand Side (RHS) = (1 – cos A)/(1 + cos A)

Therefore, LHS = RHS

(xi) (1 + sin A)/(1 – sin A) = (cosec A + 1)/(cosec A – 1)

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + sin A)/(1 – sin A)

Then consider Right Hand Side (RHS) = (cosec A + 1)/(cosec A – 1)

We know that, cosec A = 1/sin A

So,

= ((1/sin A) + 1)/((1/sin A) – 1)

= (1 + sin A)/(1 – sin A)

Therefore, LHS = RHS

(xii) cos A/(1 + sin A) = sec A – tan A

Solution:-

From the question first we consider Right Hand Side (RHS),

= sec A – tan A

We know that, sec A = 1/cos A, tan A = sin A/cos A

= (1/cos A) – (sin A/cos A)

= (1 – sin A)/cos A

= ((1 – sin A)/cos A) ((1 + sin A)/(1 + sin A))

By cross multiplication we get,

= (1 – sin2 A)/(cos A(1 + sin A))

= cos2 A/(cos A (1 + sin A))

= cos A/(1 + sin A)

Then, Left Hand Side (LHS) = cos A/(1 + sin A)

Therefore, LHS = RHS

(xiii) (tan θ + sec θ – 1)/(tan θ – sec + 1) = (1 + sin θ)/cos θ

Solution:-

From the question first we consider Left Hand Side (LHS),

= (tan θ + sec θ – 1)/(tan θ – sec + 1)

The above terms can be written as,

= (tan θ + sec θ – (sec2 θ – tan2 θ))/(1 + tan θ – sec θ)

= [tan θ + sec θ – {(sec θ + tan θ) (sec θ – tan θ)}]/(1 + tan θ – sec θ)

= [(tan θ + sec θ) {1 – (sec θ – tan θ)}]/(1 + tan θ – sec θ)

= [(tan θ + sec θ) (1 + tan θ – sec θ)]/(1 + tan θ – sec θ)

= [tan θ + sec θ]

= (1 + sin θ)/ cos θ

Then, Right Hand Side (RHS) = (1 + sin θ)/cos θ

Therefore, LHS = RHS

2. Prove the following identities:

(i) sin2 A cos2 B – cos2 A sin2 B = sin2 A – sin2 B

Solution:-

From the question first we consider Left Hand Side (LHS),

= sin2 A cos2 B – cos2 A sin2 B

We know that, cos2 B = (1 – sin2 B),

= sin2 A (1 – sin2 B) – (1 – sin2 A) sin2 B

= sin2 A – sin2 A sin2 B – sin2 B + sin2 A sin2 B

On simplification we get,

= sin2 A – sin2 B

Then, Right Hand Side (RHS) = sin2 A – sin2 B

Therefore, LHS = RHS

(ii). (1 – tan A)2 + (1 + tan A)2 = 2 sec2 A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 – tan A)2 + (1 + tan A)2

Expanding the above terms we get,

= 1 + tan2 A – 2 tan A + 1 + tan2 A + 2 tan A

= 2(1 + tan2 A)

= 2 sec2 A

Then, Right Hand Side (RHS) = 2 sec2 A

Therefore, LHS = RHS

(iii) cosec4 A – cosec2 A = cot4 A + cot4 A

Solution:-

From the question first we consider Left Hand Side (LHS),

= cosec4 A – cosec2 A

By taking common we get,

= cosec2 A (cosec2 A – 1)

Now we consider Right Hand Side (RHS) = cot4 A + cot4 A

Again taking common we get,

= cot2 A (cot2 A + 1)

We know that, cot2 A = cosec2 A – 1, cot2 A + 1 = cosec2 A

So, (cosec2 A – 1) cosec2 A

Therefore, LHS = RHS

(iv) sec2 A + cosec2 A = sec2 A cosec2 A

Solution:-

From the question first we consider Left Hand Side (LHS),

= sec2 A + cosec2 A

We know that, sec2 A = 1/cos2 A, cosec2 A = 1/sin2 A

= (1/cos2 A) + (1/sin2 A)

= (sin2 A + cos2 A)/(cos2 A sin2 A)

= 1/(cos2 A sin2)

= sec2 A cosec2 A

Then, Right Hand Side (RHS) = sec2 A cosec2 A

Therefore, LHS = RHS

(v) cos4 A – sin4 A = 2 cos2 A – 1

Solution:-

From the question first we consider Left Hand Side (LHS),

= cos4 A – sin4 A

We know that, a2 – b2 = (a + b) (a – b)

= (cos2 A – sin2 A)(cos2 A + sin2 A)

= (cos2 A – (1 – cos2 A))

= 2 cos2 A – 1

Then, Right Hand Side (RHS) = 2 cos2 A – 1

Therefore, LHS = RHS

(vii) (sec A – cos A) (sec A + cos A) = sin2 A + tan2 A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (sec A – cos A) (sec A + cos A)

We know that, a2 – b2 = (a + b) (a – b)

= (sec2 A – cos2 A)

Also we know that, sec2 A = 1 + tan2 A, cos2 A = 1 – sin2 A

= 1 + tan2 A – (1 – sin2 A)

= tan2 A + sin2 A

Then, Right Hand Side (RHS) = tan2 A + sin2 A

Therefore, LHS = RHS

(viii) (cos A + sin A)2 + (cos A – sin A)2 = 2

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cos A + sin A)2 + (cos A – sin A)2

We know that, (a + b)2 = a2 + 2ab + b2, (a – b)2 = a2 – 2ab + b2

= cos2 A + sin2 A + 2 cos A sin A + cos2 A + sin2 A – 2 cos A sin A

By simplification we get,

= 2(cos2 A + sin2 A)

= 2

Then, Right Hand Side (RHS) = 2

Therefore, LHS = RHS

(ix) (cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cosec A – sin A) (sec A – cos A) (tan A + cot A)

We know that, cosec A = 1/sin A, sec A = 1/cos A, cot A = 1/tan A

= ((1/sin A) – sin A) ((1/cos A) – cos A) (tan A + (1/tan A))

= ((1 – sin2 A)/sin A) ((1 – cos2 A)/cos A) ((sin A/cos A) + (cos A/sin A))

= (cos2 A/sin A) (sin2 A/cos A) ((sin2 A + cos2 A)/(sin A cos A))

By simplification we get,

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(x) sec2 A cosec2 A = tan2 A + cot2 A + 2

Solution:-

From the question first we consider Left Hand Side (LHS),

= sec2 A cosec2 A

We know that, sec2 A = 1/cos2 A, cosec2 A = 1/sin2 A

= 1/(cos2 A sin2 A)

Now consider RHS = tan2 A + cot2 A + 2

= tan2 A + cot2 A + 2 tan2 A cot2 A

= (tan A + cot A)2

= ((sin A/cos A) + (cos A/sin A))2

= ((sin2 A + cos2 A)/(sin A cos A))

= 1/(cos2 A sin2 A)

Therefore, LHS = RHS

(xii) (cos A/(1 – tan A)) + (sin A/(1 – cot A)) = sin A + cos A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cos A/(1 – tan A)) + (sin A/(1 – cot A))

= (cos A/(1 – (sin A/cos A))) + (sin A/(1 – (cos A/sin A)))

= (cos A/((cos A – sin A)/cos A)) + (sin A/((sin A – cos A)/sin A))

= (cos2A/(cos A – sin A)) + (sin2 A/(sin A – cos A))

= (cos2 A – sin2 A)/(cos A – sin A)

= sin A + cos A

Then, Right Hand Side (RHS) = sin A + cos A

Therefore, LHS = RHS

(xiii) (cosec A – sin A) (sec A – cos A) = 1/(tan A + cot A)

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cosec A – sin A) (sec A – cos A)

We know that, cosec A = 1/sin A, sec A = 1/cos A

= ((1/sin A) – sin A) ((1/cos) – cos A)

= ((1 – sin2 A)/sin A)((1 – cos2 A)/cos A)

= (cos2 A/sin A) (sin2 A/cos A)

= cos A sin A

Now consider the RHS = 1/(tan A + cot A)

= (1/((sin A/cos A)+(cos A/sin A))

= 1/((sin2 A + cos2 A)/(sin A cos A))

= cos A sin A

Therefore, LHS = RHS

(xiv) sin4 A + cos4 A = 1 – 2 sin2 A cos2 A

Solution:-

From the question first we consider Left Hand Side (LHS),

= sin4 A + cos4 A

= 1 – 2 sin2 A cos2 A

= sin4 A + cos4 A + 2 sin2 A cos2 A

= (sin2A)2+(cos2A)2+ 2 sin2A cos2A − 2sin2Acos2A   

[Adding and subtracting 2 sin2A cos2A]

= (sin2A + cos2A)2 – 2 sin2 A cos2 A

= 1 – 2 sin2 A cos2 A

Then, Right Hand Side (RHS) = 1 – 2 sin2 A cos2 A

Therefore, LHS = RHS

3. Prove the following identities:

(i) (sin A/(1 + cos A)) + ((1 + cos A)/sin A) = 2 cosec A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (sin A/(1 + cos A)) + ((1 + cos A)/sin A)

By cross multiplication we get,

= ((1 + cos A)2 + sin2 A)/(sin A (1 + cos))

We know that, (a + b)2 = a2 + 2ab + b2

Then,

= (1 + 2 cos A + cos2 A + sin2 A)/(sin A (1 + cos A)

= (2 + 2 cos A)/(sin A (1 + cos A))

Now taking common outside we get,

= (2(1 + cos A))/(sin A (1 + cos A))

= 2 cosec A

Then, Right Hand Side (RHS) = 2 cosec A

Therefore, LHS = RHS

(ii) (1 + cos A)/(1 – cos A) = (cosec A + cot A)2

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + cos A)/(1 – cos A)

Now, multiply and divide by (1 + cos A) we get,

= ((1 + cos A)/(1 – cos A)) ((1 + cos A)/(1 + cos A))

By cross multiplication we get,

= (1 + cos A)2/(1 – cos2 A)

We know that, sin2 A + cos2 A = 1

So, (1 + cos A)2/sin2 A

= ((1 + cos A)/sin A)2

= ((1/sin A) + (cos A/sin A))2

= (cosec A + cot A)2

Then, Right Hand Side (RHS) = (cosec A + cot A)2

Therefore, LHS = RHS

(iii) (cot A + tan B)/(cot B + tan A) = cot A tan B

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cot A + tan B)/(cot B + tan A)

We know that, cot A = 1/tan A

= ((1/tan A) + (tan B))/((1/tan A) + tan A)

= ((1 + tan A tan B)/tan A)/((1 + tan A tan B)/tan B)

= ((1 + tan A tan B)/tan A) (tan B/(1 + tan A tan B))

= tan B/tan A

= (1/tan A) tan B

= cot A tan B

Then, Right Hand Side (RHS) = cot A tan B

Therefore, LHS = RHS

(iv) 1/(tan A + cot A) = sin A cos A

Solution:-

From the question first we consider Left Hand Side (LHS),

= 1/(tan A + cot A)

We know that, tan A = sin A/cos A, cot A = cos A/sin A

= 1/((sin A/cos A) + (cos A/sin A))

By cross multiplication we get,

= 1/((sin2 A + cos2 A)/(sin A cos A))

Also we know that, sin2 A + cos2 A = 1

= 1/(1/(sin A cos A))

= sin A cos A

Then, Right Hand Side (RHS) = sin A cos A

Therefore, LHS = RHS

(v) tan A – cot A = (1 – 2 cos2 A)/(sin A cos A)

Solution:-

From the question first we consider Left Hand Side (LHS),

= tan A – cot A

We know that, tan A = sin A/cos A, cot A = cos A/sin A

Then,

= (sin A/cos A) – (cos A /sin A)

= (sin2 A – cos2 A)/(sin A cos A)

We know that, sin2 A = 1 – cos2 A

= (1 – cos2 A – cos2 A)/(sin A cos A)

So,

= (1 – 2 cos2 A)/(sin A cos A)

Then, Right Hand Side (RHS) = (1 – 2 cos2 A)/(sin A cos A)

Therefore, LHS = RHS

(vi) ((1 + tan2 A) cot A)/cosec2 A = tan A

Solution:-

From the question first we consider Left Hand Side (LHS),

= ((1 + tan2 A) cot A)/cosec2 A

We know that, 1 + tan2 A = sec2

= (sec2 cot A)/cosec2 A

Also we know that, sec2 A = 1/cos2 A, cot A = cos A/sin A

= ((1/cos2 A)(cos A/sin A))/(1/(sin2 A))

= (1/(cos A sin A))/(1/sin2 A)

= sin A/cos A

= tan A

Then, Right Hand Side (RHS) = tan A

Therefore, LHS = RHS

(vii) cosec A + cot A = (1/(cosec A – cot A))

Solution:-

From the question first we consider Left Hand Side (LHS),

= cosec A + cot A

Now multiply and divide by cosec A – cot A we get,

= ((cosec A + cot A)/1) ((cosec A – cot A)/(cosec A – cot A))

By cross multiplication we get,

= (cosec2 A – cot2 A)/(cosec A – cot A)

= (1 + cot2 A – cot2 A)/(cosec A – cot A)

= (1/(cosec A – cot A))

Then, Right Hand Side (RHS) = (1/(cosec A – cot A))

Therefore, LHS = RHS

(viii) ((cosec A)/(cosec A – 1)) + ((cosec A)/(cosec A + 1)) = 2 sec2 A

Solution:-

From the question first we consider Left Hand Side (LHS),

= ((cosec A)/(cosec A – 1)) + ((cosec A)/(cosec A + 1))

By cross multiplication we get,

= (cosec2 A + cosec A + cosec2 A – cosec A)/(cosec2 A – 1)

We know that, cosec2 A – 1 = cot2 A

= 2 cosec2 A/cot2 A

Also we know that, cosec2 A = 1/sin2 A

= (2/sin2 A)/(cos2 A/sin2 A)

= 2/cos2 A

= 2 sec2 A

Then, Right Hand Side (RHS) = 2 sec2 A

Therefore, LHS = RHS

(ix) (1 + cos A)/(1 – cos A) = (tan2 A)/(sec A – 1)2

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + cos A)/(1 – cos A)

We know that, cos A = 1/sec A

Then,

= (1 + (1/sec A))/(1 – (1/sec))

= (sec A + 1)/(sec A – 1)

Now multiply and divide by (sec A – 1) we get,

= ((sec A + 1)/(sec A – 1)) ((sec A – 1)/(sec A – 1))

By cross multiplication we get,

= (sec2 A – 1)/(sec A – 1)2

= tan2 A/(sec A – 1)2

Then, Right Hand Side (RHS) = tan2 A/(sec A – 1)2

Therefore, LHS = RHS

(x) (cot A – cosec A)2 = ((1 – cos A)/(1 + cos A))

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cot A – cosec A)2

We know that, cot A = cos A/sin A, cosec A = 1/sin A

= ((cos A/sin A) – (1/sin A))2

= ((cos A – 1)/sin A)2

= (cos A – 1)2/sin2 A

= (cos A – 1)2/1 – cos2 A

= (-(1 – cos A)2/((1 – cos A) (1 + cos A))

= ((1 – cos A) (1 – cos A))/((1 – cos A)(1 + cos A))

= (1 – cos A)/(1 + cos A)

Then, Right Hand Side (RHS) = (1 – cos A)/(1 + cos A)

Therefore, LHS = RHS

4. Prove the following identities:

(i) √(cosec2 q – 1) = cos q cosec q

Solution:-

From the question first we consider Left Hand Side (LHS),

= √(cosec2 q – 1)

We know that, cosec2 q – 1 = cot2 q

= √(cot2 q)

Then,

= cot q

= cos q/sin q

= cos q (1/sin q)

Also we know that, 1/ sin q = cosec q

= cos q cosec q

Then, Right Hand Side (RHS) = cos q cosec q

Therefore, LHS = RHS

(ii) √((1 + sin q)/(1 – sin q)) + √((1 – sin q)/(1 + sin q)) = 2 sec q

Solution:-

From the question first we consider Left Hand Side (LHS),

= √((1 + sin q)/(1 – sin q)) + √((1 – sin q)/(1 + sin q))

Then,

= √(((1 + sin q)/(1 – sin q)) ((1 + sin q)/(1 + sin q))) + √(((1 – sin q)/(1 + sin q)) ((1 + sin q)/(1 + sin q)))

= √((1 + sin q)2/(1 – sin2 q)) + √((1 – sin q)2/(1 – sin2 q))

We know that, 1 – sin2 q = cos2 q

= √((1 + sin q)2/cos2 q) + √((1 – sin q)2/cos2 q)

= ((1 + sin q)/cos q) + ((1 – sin q)/cos q)

= 2/cos q

= 2 sec q

Then, Right Hand Side (RHS) = 2 sec q

Therefore, LHS = RHS

(iii) √((1 – cos A)/(1 + cos A)) = (sin A/(1 + cos A))

Solution:-

From the question first we consider Left Hand Side (LHS),

= √((1 – cos A)/(1 + cos A))

Then,

= √(((1 – cos A)/(1 + cos A)) ((1 + cos A)/(1 + cos A)))

= √((1 – cos2 A)/(1 + cos A)2)

= √((sin2 A)/(1 + cos A)2)

= sin A/(1 + cos A)

Then, Right Hand Side (RHS) = sin A/(1 + cos A)

Therefore, LHS = RHS

(iv) √((1 + cos A)/(1 – cos A)) = cosec A + cot A

Solution:-

From the question first we consider Left Hand Side (LHS),

= √((1 + cos A)/(1 – cos A))

Then,

= √(((1 + cos A)/(1 – cos A)) ((1 + cos A)/(1 + cos A)))

= √((1 + cos A)2/(1 – cos2 A))

We know that, 1 – cos2 A = sin2 A

= √((1 + cos2 A)/(sin2 A))

= √((1 + cos A)/(sin A))2

= √((1/sin A) + (cos A/sin A))2

= √(cosec A + cot A)2

= cosec A + cot A

Then, Right Hand Side (RHS) = cosec A + cot A

Therefore, LHS = RHS

(v) √((sec q – 1)/(sec q + 1)) + √((sec q + 1)/(sec q – 1)) = 2 cosec q

Solution:-

From the question first we consider Left Hand Side (LHS),

= √((sec q – 1)/(sec q + 1)) + √((sec q + 1)/(sec q – 1))

Then,

= √(((sec q – 1)/(sec q + 1)) ((sec q – 1)/(sec q – 1))) + √(((sec q + 1)/(sec q – 1)) ((sec q – 1)/(sec q – 1)))

By simplification we get,

= √((sec q – 1)2/(sec2 q – 1)) + √((sec q + 1)2/(sec2 q – 1))

We know that, sec2 q – 1 = tan2 q

= √((sec q – 1)2/tan2 q) + √((sec q + 1)2/tan2 q)

= ((sec q – 1)/tan q) + ((sec q + 1)/tan q)

= (sec q – 1 + sec q + 1)/tan q

= (2 sec q/tan q)

= (2/cos q)/(sin q/cos q)

= 2/sin q

= 2 cosec q

Then, Right Hand Side (RHS) = 2 cosec q

Therefore, LHS = RHS

5. Prove the following identities:

(i) (sec θ – tan θ)2 = (1 – sin θ)/(1 + sin θ)

Solution:-

From the question first we consider Left Hand Side (LHS),

= (sec θ – tan θ)2

We know that, sec θ = 1/cos θ, tan θ = sin θ/cos θ

= ((1/cos θ) – (sin θ/cos θ))2

By taking LCM we get,

= ((1 – sin θ)/cos θ)2

= (1 – sin θ)2/cos2 θ

Also we know that, cos2 θ = 1 – sin2 θ

= (1 – sin θ)2/(1 – sin2 θ)

= (1 – sin θ)2/((1 – sin θ)(1 + sin θ))

= (1 – sin θ)/(1 + sin θ)

Then, Right Hand Side (1 – sin θ)/(1 + sin θ)

Therefore, LHS = RHS

(ii) (1/(sin A + cos A)) + (1/(sin A – cos A)) = 2 sin A/(1 – 2 cos2 A)

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1/(sin A + cos A)) + (1/(sin A – cos A))

By taking LCM we get,

= (sin A – cos A + sin A + cos A)/(sin2 A – cos2 A)

= 2 sin A/(1 – cos2 A – cos2 A)

= 2 sin A/(1 – 2 cos2 A)

Then, Right Hand Side 2 sin A/(1 – 2 cos2 A)

Therefore, LHS = RHS

(iii) ((sin A + cos A)/(sin A – cos A)) + ((sin A – cos A)/(sin A + cos A)) = 2 /(2 sin2 A – 1)

Solution:-

From the question first we consider Left Hand Side (LHS),

= ((sin A + cos A)/(sin A – cos A)) + ((sin A – cos A)/(sin A + cos A))

By taking LCM we get,

= ((sin A + cos A)2 + (sin A – cos A)2)/((sin A + cos A)(sin A – cos A))

Then,

=(sin2 A + cos2 A + 2 sin A cos A + sin2 A + cos2 A – 2 sin cos A)/(sin2 A – cos2 A)

= (2 sin2 A + 2 cos2 A + 2 sin A cos A – 2 sin cos A)/(sin2 A – cos2 A)

= (2 sin2 A + 2 cos2 A)/(sin2 A – cos2 A)

= 2(sin2 A + cos2 A)/(sin2 A – cos2 A)

We know that, sin2 A + cos2 A = 1

= 2(1)/(sin2 A – cos2 A)

= 2/(sin2 A – cos2 A)

= 2/(sin2 A – (1 – sin2 A))

= 2/(2 sin2 A – 1)

Then, Right Hand Side 2/(2 sin2 A – 1)

Therefore, LHS = RHS

(iv) tan2 A – tan2 B = (sin2A – sin2 B)/(cos2 A cos2 B)

Solution:-

From the question first we consider Left Hand Side (LHS),

= tan2 A – tan2 B

We know that, tan2 θ = sin2 θ /cos2 θ,

= (sin2 A cos2 B – cos2 A sin2 B)/(cos2 A cos2 B)

Then,

= (((1 – cos2 A) cos2 B) – (cos2 A sin2 B))/(cos2 A cos2 B)

= (cos2 B – cos2 A cos2 B – cos2 A + cos2 A cos2 B)/(cos2 A cos2 B)

By simplification we get,

= (cos2 B – cos2 A)/(cos2 A cos2 B)

= ((1 – sin2 B) – (1 – sin2 A))/(cos2 A cos2 B)

= (sin2 A – sin2 B)/(cos2 A cos2 B)

Then, Right Hand Side (sin2 A – sin2 B)/(cos2 A cos2 B)

Therefore, LHS = RHS

(v) (cos A/(1 – tan A)) + (sin2 A/(sin A – cos A)) = cos A + sin A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cos A/(1 – tan A)) + (sin2 A/(sin A – cos A))

We know that, tan A = sin A/cos A

= (cos A/(1 – (sin A/cos A))) + (sin2 A/(sin A – cos A))

Then,

= (cos A/((cos A – sin A)/cos A)) + (sin2 A/(sin A – cos A))

= (cos2 A/(cos A – sin A)) – (sin2 A/(cos A – sin A))

= (cos2 A – sin2 A)/(cos A – sin A)

= ((cos A + sin A)(cos A – sin A))/(cos A – sin A)

By simplification we get,

= cos A + sin A

Then, Right Hand Side cos A + sin A

Therefore, LHS = RHS

(vi) (1 + tan2 A) + (1 + (1/tan2 A)) = (1/(sin2 A – sin4 A))

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + tan2 A) + (1 + (1/tan2 A))

We know that, tan2 A = sin2 A/cos2 A

= (1 + (sin2 A/cos2 A)) + (1 + (1/(sin2A/cos2 A)))

By taking LCM we get,

= ((cos2 A + sin2 A)/cos2 A) + ((cos2 A + sin2 A)/sin2 A)

Also we know that, cos2 A + sin2 A = 1

So,

= (1/(1 – sin2 A)) + (1/sin2 A)

= (sin2 A + 1 – sin2 A)/(sin2 A(1 – sin2 A))

= 1/(sin2 A – sin4 A)

Then, Right Hand Side = 1/(sin2 A – sin4 A)

Therefore, LHS = RHS

(vii) ((cos3 A + sin3 A)/(cos A + sin A)) + ((cos3 A – sin3 A)/(cos A – sin A)) = 2

Solution:-

From the question first we consider Left Hand Side (LHS),

= ((cos3 A + sin3 A)/(cos A + sin A)) + ((cos3 A – sin3 A)/(cos A – sin A))

By taking LCM we get,

= ((cos3 A + sin3 A)(cos A – sin A) + (cos3 A – sin3 A)(cos A + sin A))/(cos2 A – sin2 A)

= 2(cos4 A – sin4 A)/(cos2 A – sin2 A)

= (2(cos2 A + sin2 A)(cos2 A – sin2 A))/(cos2 A – sin2 A)

= 2(cos2 A + sin2 A)

We know that, cos2 A + sin2 A = 1

= 2

Then, Right Hand Side = 2

Therefore, LHS = RHS

(viii) (tan θ + (1/cos θ))2 + (tan θ – (1/cos θ))2 = 2((1 + sin2 θ)/(1 – sin2 θ))

Solution:-

From the question first we consider Left Hand Side (LHS),

= (tan θ + (1/cos θ))2 + (tan θ – (1/cos θ))2

We know that, tan θ = sin θ/cos θ

= ((sin θ/cos θ) + (1/cos θ))2 + ((sin θ/cos θ) – (1/cos θ))2

= ((sin θ + 1)/cos θ)2 + ((sin θ – 1)/cos θ)2

= ((sin θ + 1)2/cos2 θ) + ((sin θ – 1)/cos2 θ)

= ((sin θ + 1)2 + (sin θ – 1)2)/cos2 θ

Also we know that, (a + b)2 = a2 + 2ab + b2

= (sin2 θ + 1 + 2 sin θ + sin2 θ + 1 – 2 sin θ)/(1 – sin2 θ)

= 2(1 + sin2 θ)/(1 – sin2 θ)

Then, Right Hand Side = 2(1 + sin2 θ)/(1 – sin2 θ)

Therefore, LHS = RHS

(ix) ((sin A – sin B)/(cos A + cos B)) + ((cos A – cos B)/(sin A + sin B)) = 0

Solution:-

From the question first we consider Left Hand Side (LHS),

= ((sin A – sin B)/(cos A + cos B)) + ((cos A – cos B)/(sin A + sin B))

By taking LCM we get,

= (((sin A + sin B)(sin A – sin B)) + ((cos A + cos B)(cos A – cos B)))/((cos A + cos B)(sin A – sin B))

By simplification we get,

= ((sin2 A – sin2 B) + (cos2 A – cos2 B))/((cos A + cos B)(sin A – sin B))

= ((sin2 A + cos2 A) – (sin2 B + cos2 B))/((cos A + cos B)(sin A – sin B))

We know that, sin2 A + cos2 A = 1

= (1 – 1)/ ((cos A + cos B)(sin A – sin B))

= 0/((cos A + cos B)(sin A – sin B))

= 0

Then, Right Hand Side = 0

Therefore, LHS = RHS

(x) (1/(cos A + sin A – 1)) + (1/(cos A + sin A + 1)) = cosec A + sec A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1/(cos A + sin A – 1)) + (1/(cos A + sin A + 1))

By taking LCM we get,

= (cos A + sin A + 1 + cos A + sin A – 1)/((cos A + sin A)2 – 1)

We know that, (a + b)2 = a2 + 2ab + b2

= (2(cos A + sin A))/(cos2 A + sin2 A + 2 cos A sin A – 1)

= (cos A + sin A)/(cos A sin A)

= (cos A/(cos A sin A)) + (sin A/(cos A sin A))

= (1/sin A) + (1/cos A)

= cosec A + sec A

Then, Right Hand Side = cosec A + sec A

Therefore, LHS = RHS

(xi) (cot A + cosec A – 1)/(cot A – cosec A + 1) = (cos A + 1)/sin A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cot A + cosec A – 1)/(cot A – cosec A + 1)

We know that, cosec2 A – cot2 A = 1

= (cot A + cosec – (cosec2 A – cot2 A))/(cot A – cosec A + 1)

Also we know that, (a2 – b2) = (a + b) (a – b)

= [cot A + cosec A – ((cosec A – cot A)(cosec A + cot A))]/(cot A – cosec A + 1)

= (cot A + cosec A [1 – cosec A + cot A])/(cot A – cosec A + 1)

= cot A + cosec A

= (cos A/sin A) + (1/sin A)

= (1 + cos A)/sin A

Then, Right Hand Side = (1 + cos A)/sin A

Therefore, LHS = RHS

(xii) (sec A – 1)/(sec A + 1) = (sin2 A)/(1 + cos A)2

Solution:-

From the question first we consider Left Hand Side (LHS),

= (sec A – 1)/(sec A + 1)

We know that, sec A = 1/cos A

= ((1/cos A) – 1)/((1/cos) + 1)

= (1 – cos A)/(1 + cos A)

Then,

= ((1 – cos A)/(1 + cos A)) × ((1 + cos A)/(1 + cos A))

By simplification we get,

= (1 – cos2 A)/(1 + cos A)2

Also we know that, 1 – cos2 A = sin2 A

= sin2 A/(1 + cos A)2

Then, Right Hand Side = sin2 A/(1 + cos A)2

Therefore, LHS = RHS

6. Prove the following identities:

(i) (1 + cot A)2 + (1 – cot A)2 = 2 cosec2 A

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + cot A)2 + (1 – cot A)2

We know that, (a + b)2 = a2 + 2ab + b2

= 1 + cot2 A + 2 cot A + 1 + cot2 A – 2 cot A

= 2 + 2 cot2 A

Taking common terms outside we get,

= 2 (1 + cot2 A)

Also we know that, 1 + cot2 A = cosec2 A

= 2 cosec2 A

Then, Right Hand Side = 2 cosec2 A

Therefore, LHS = RHS

(ii) (cosec θ/(tan θ + cot θ)) = cos θ

Solution:-

From the question first we consider Left Hand Side (LHS),

= (cosec θ/(tan θ + cot θ))

We know that, cosec θ = 1/sin θ, tan θ = sin θ/cos θ, cot θ = cos θ/sin θ

Then,

= (1/sin θ)/((sin θ/cos θ) + (cos θ/sin θ))

Taking LCM in the denominator we get,

= (1/sin θ)/((sin2 θ + cos2 θ)/(cos θ sin θ))

Also we know that, sin2 θ + cos2 θ = 1

= (1/sin θ)/(1/cos θ sin θ)

= (1/sin θ) × ((cos θ sin θ)/1)

= cos θ

Then, Right Hand Side = cos θ

Therefore, LHS = RHS

(iii) (1 + tan2 θ) sin θ cos θ = tan θ

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + tan2 θ) sin θ cos θ

We know that, tan2 θ = sin2 θ/cos2 θ

= (1 + (sin2 θ/cos2 θ)) sin θ cos θ

Taking LCM we get,

= ((cos2 θ + sin2 θ)/cos2 θ) sin θ cos θ

Also we know that, sin2 θ + cos2 θ = 1

= (1/cos2 θ) sin θ cos θ

= sin θ/cos θ

= tan θ

Then, Right Hand Side = tan θ

Therefore, LHS = RHS

(iv) ((1 + sin θ)/(cosec θ – cot θ)) – ((1 – sin θ)/(cosec θ + cot θ)) = 2 (1 + cot θ)

Solution:-

From the question first we consider Left Hand Side (LHS),

= ((1 + sin θ)/(cosec θ – cot θ)) – ((1 – sin θ)/(cosec θ + cot θ))

By taking LCM we get,

= [((1 + sin θ)(cosec θ + cot θ)) – ((1 – sin θ)(cosec θ – cot θ))]/(cosec2 θ – cot2 θ)

We know that, 1 + cot2 θ = cosec2 θ

= (cosec θ + cot θ + 1 + cos θ – cosec θ + cot θ + 1 – cos θ)/(1 + cot2 θ – cot2 θ)

By simplification we get,

= 2 + 2 cot θ

Taking common terms outside we get,

= 2(1 + cot θ)

Then, Right Hand Side = 2(1 + cot θ)

Therefore, LHS = RHS

(v) (1 + cot A + tan A) (sin A – cos A) = (sec A/cosec2 A) – (cosec A/sec2 A)

Solution:-

From the question first we consider Left Hand Side (LHS),

= (1 + cot A + tan A) (sin A – cos A)

We know that, cot A = cos A/sin A, tan A = sin A/cos A

= [1 + (cos A/sin A) + (sin A/cos A)] (sin A – cos A)

By taking LCM we get,

= [(sin A cos A + cos2 A + sin2 A)/(sin A cos A)] (sin A – cos A)

Also we know that, (a3 – b3) = (a – b) (a2 + b2 + ab)

So,

= (sin3 A – cos3 A)/(sin A cos A)

= (sin3 A/(sin A cos A)) – (cos3 A/(sin A cos A))

= (sin2 A/cos A) – (cos2 A/sin A)

= ((1/cos A) × sin2 A) – ((1/sin A) × cos2 A)

= sec A sin2 A – cosec A cos2 A

= (sec A/cosec2 A) – (cosec A/sec2 A)

Then, Right Hand Side = (sec A/cosec2 A) – (cosec A/sec2 A)

Therefore, LHS = RHS

(vi) 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1 = 0

Solution:-

From the question first we consider Left Hand Side (LHS),

= 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ) + 1

The above terms can be written as,

= 2[(sin2 θ)3 + (cos2 θ)3] – 3 (sin4 θ + cos4 θ) + 1

We know that, (a3 + b3) = (a + b) (a2 + b2 – ab)

= 2[(sin2 θ + cos2 θ) ((sin2 θ) + (cos2 θ)2 – sin2 θ cos2 θ)] – 3(sin4 θ + cos4 θ) + 1

= 2[(sin2 θ)2 + (cos2 θ)2 – sin2 θ cos2 θ] – 3 (sin4 θ + cos4 θ) + 1

= 2 sin4 θ + 2 cos4 θ – 2 sin2 θ cos2 θ – 3 sin4 θ – 3 cos4 θ + 1

= – sin4 θ – cos4 θ – 2 sin2 θ cos2 θ + 1

= -(sin4 θ + cos4 θ + 2 sin2 θ cos2 θ) + 1

= – (sin2 θ + cos2 θ)2 + 1

Also we know that, sin2 θ + cos2 θ = 1

= – 1 + 1

= 0

Then, Right Hand Side = 0

Therefore, LHS = RHS

(vii) sin8 θ – cos8 θ = (sin2 θ – cos2 θ)(1 – 2 sin2 θ cos2 θ)

Solution:-

From the question first we consider Left Hand Side (LHS),

= sin8 θ – cos8 θ

The above terms can be written as,

= (sin4 θ)2 – (cos4 θ)2

We know that, (a2 – b2) = (a + b)(a – b)

= (sin4 θ – cos4 θ) (sin4 θ + cos4 θ)

= ((sin2 θ)2 – (cos2 θ)2) (sin4 θ + cos4 θ)

= (sin2 θ – cos2 θ) (sin2 θ + cos2 θ)(sin4 θ + cos4 θ)

= (sin2 θ – cos2 θ)(sin4 θ + cos4 θ)

= (sin2 θ – cos2 θ)((sin2 θ)2 + (cos2 θ)2 + 2sin2 θ cos2 θ – 2sin2 θ cos2 θ)

= (sin2 θ – cos2 θ)((sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ)

= (sin2 θ – cos2 θ)(1 – 2sin2 θ cos2 θ)

Then, Right Hand Side = (sin2 θ – cos2 θ)(1 – 2sin2 θ cos2 θ)

Therefore, LHS = RHS

(viii) sec4 A – sec2 A = (sin2 A/cos4 A)

Solution:-

From the question first we consider Left Hand Side (LHS),

= sec4 A – sec2 A

We know that, sec A = 1/cos A

= (1/cos4 A) – (1/cos2 A)

= (1 – cos2 A)/cos4 A

Also we know that, 1 – cos2 A = sin2 A

= sin2 A/cos4 A

Then, Right Hand Side = sin2 A/cos4 A

Therefore, LHS = RHS

(ix) (tan2 θ/(tan2 θ – 1)) + (cosec2 θ/(sec2 θ – cosec2 θ)) = (1/(sin2 θ – cos2 θ))

Solution:-

From the question first we consider Left Hand Side (LHS),

= (tan2 θ/(tan2 θ – 1)) + (cosec2 θ/(sec2 θ – cosec2 θ))

We know that, tan2 θ = sin2 θ/cos2 θ, cosec2 θ = 1/sin2 θ, sec2 θ = 1/cos2 θ

= ((sin2 θ/cos2 θ)/((sin2 θ/cos2 θ) – 1)) + ((1/sin2 θ)/((1/cos2 θ) – (1/sin2 θ)))

= (sin2 θ/(sin2 θ – cos2 θ)) + ((1/sin2 θ)/((sin2 θ – cos2 θ)/(cos2 θ sin2 θ)))

= (sin2 θ/(sin2 θ – cos2 θ)) + (cos2 θ/(sin2 θ – cos2 θ))

= (sin2 θ + cos2 θ)/(sin2 θ – cos2 θ)

= 1/(sin2 θ – cos2 θ)

Then, Right Hand Side = 1/(sin2 θ – cos2 θ)

Therefore, LHS = RHS

(x) (sec2 θ – sin2 θ)/(tan2 θ) = cosec2 θ – cos2 θ

Solution:-

From the question first we consider Left Hand Side (LHS),

= (sec2 θ – sin2 θ)/(tan2 θ)

We know that, sec2 θ = 1/cos2 θ, tan2 θ = sin2 θ/cos2 θ

= ((1/cos2 θ) – sin2 θ)/(sin2 θ/cos2 θ)

= ((1 – sin2 θ cos2 θ)/cos2 θ)/(sin2 θ/cos2 θ)

= (1 – sin2 θ cos2 θ)/sin2 θ

= (1/sin2 θ) – ((sin2 θ cos2 θ)/sin2 θ)

= cosec2 θ – cos2 θ

Then, Right Hand Side = cosec2 θ – cos2 θ

Therefore, LHS = RHS

(xi) ((cos3 θ + sin3 θ)/(cos θ + sin θ)) + ((cos3 θ – sin3 θ)/(cos θ – sin θ)) = 2

Solution:-

From the question first we consider Left Hand Side (LHS),

= ((cos3 θ + sin3 θ)/(cos θ + sin θ)) + ((cos3 θ – sin3 θ)/(cos θ – sin θ))

By taking LCM we get,

= [((cos3 θ + sin3 θ)(cos θ – sin θ)) + ((cos3 θ – sin3 θ)(cos θ + sin θ))]/((cos θ + sin θ)(cos θ – sin θ))

Then,

= (cos4 θ – cos3 θ sin θ + sin3 θ cos θ – sin4 θ + cos4 θ + cos3 θ sin θ – sin3 θ cos θ – sin4 θ)/(cos2 θ – sin2 θ)

By simplification we get,

= (2 cos4 θ – 2 sin4 θ)/(cos2 θ – sin2 θ)

Taking common outside,

= 2(cos4 θ – sin4 θ)/(cos2 θ – sin2 θ)

= (2(cos2 θ + sin2 θ)(cos2 θ – sin2 θ))/(cos2 θ – sin2 θ)

= 2(cos2 θ + sin2 θ)

We know that, sin2 θ + cos2 θ = 1

= 2

Then, Right Hand Side = 2

Therefore, LHS = RHS

(xii) (tan θ + sin θ)/(tan θ – sin θ) = (sec θ + 1)/(sec θ – 1)

Solution:-

From the question first we consider Left Hand Side (LHS),

= (tan θ + sin θ)/(tan θ – sin θ)

We know that, tan θ = sin θ/cos θ

= ((sin θ/cos θ) + sin θ)/((sin θ/cos θ) – sin θ)

= (sin θ + sin θ cos θ)/(sin θ – sin θ cos θ)

= (sin θ(1 + cos θ))/(sin θ(1 – cos θ))

= (1 + cos θ)/(1 – cos θ)

Also we know that, cos θ = 1/sec θ

= (1 + (1/cos θ))/(1 – (1/sec θ))

= ((sec θ + 1)/sec θ)/((sec θ – 1)sec θ)

= (sec θ + 1)/(sec θ – 1)

Then, Right Hand Side = (sec θ + 1)/(sec θ – 1)

Therefore, LHS = RHS

7. If m = a sec A + b tan A and n = a tan A + b sec A, prove that m2 – n2 = a2 – b2.

Solution:-

From the question it is given that,

m = a sec A + b tan A

n = a tan A + b sec A

We have to prove that, m2 – n2 = a2 – b2

Then,

m2 – n2 = (a sec A + b tan A)2 – (a tan A + b sec A)2

We know that, (a + b)2 = a2 + b2 + 2ab

= a2sec2A + b2 tan2 A + 2a sec A b tan A – (a2 tan2 A + b2 sec2 A + 2ab sec A tan A)

= sec2 A(a2 – b2) + tan2 A(b2 – a2)

= (a2 – b2)[sec2 A – tan2 A]

Also we know that, sec2 A – tan2 A = 1

= (a2 – b2)

Hence it is proved that, m2 – n2 = a2 – b2

8. If x = r sin A cos B, y = r sin A sin B and z = r cos A, prove that x2 + y2 + z2 = r2.

Solution:-

From the question it is given that,

x = r sin A cos B

y = r sin A sin B

z = r cos A

We have to prove that, x2 + y2 + z2 = r2

First we consider Left Hand Side (LHS),

= x2 + y2 + z2

= (r sin A cos B)2 + (r sin A sin B)2 + (r cos A)2

= r2 sin2 A cos2 B + r2 sin2 A sin2 B + r2 cos2 A

Taking common terms outside we get,

= r2 sin2 A (cos2 B + sin2 B) + r2 cos2 A

= r2 (sin2 A + cos2 A)

We know that, sin2 A + cos2 A = 1

= r2

Then, Right Hand Side = r2

Therefore, LHS = RHS

Hence it is proved that, x2 + y2 + z2 = r2

9. If sin A + cos A = m and sec A + cosec A = n, prove that n (m2 – 1) = 2m

Solution:-

From the question it is given that,

sin A + cos A = m

sec A + cosec A = n

We have to prove that, n (m2 – 1) = 2m

First we consider Left Hand Side (LHS),

= n(m2 – 1)

= (sec A + cosec A)(( sin A + cos A)2 – 1)

We know that, sec A = 1/cos A, cosec A = 1/sin A

= ((1/cos A) + (1/sin A))[sin2 A + cos2 A + 2 sin A cos A – 1]

= ((cos A + sin A)/(sin A cos A)) (1 + 2 sin A cos A – 1)

= ((cos A + sin A)/(sin A cos A))(2 sin A cos A)

= 2(sin A + cos A)

= 2m

Then, Right Hand Side = 2m

Therefore, LHS = RHS

Hence it is proved that, n (m2 – 1) = 2m

10. If x = a cos θ, y = b cot θ, prove that (a2/x2) – (b2/y2) = 1.

Solution:-

From the question it is given that,

x = a cos θ

y = b cot θ

We have to prove that, (a2/x2) – (b2/y2) = 1

First we consider Left Hand Side (LHS),

= (a2/x2) – (b2/y2)

= (a2/a2cos2 θ) – (b2/b2 cot2 θ)

= (1/cos2 θ) – (1/cot2 θ)

= sec2 θ – tan2 θ

We know that, 1 + tan2 θ = sec2 θ

= 1

Then, Right Hand Side = 1

Therefore, LHS = RHS

Hence it is proved that, (a2/x2) – (b2/y2) = 1.

11. If sec θ + tan θ = m, sec θ – tan θ = n, prove that mn = 1.

Solution:-

From the question it is given that,

sec θ + tan θ = m

sec θ – tan θ = n

We have to prove that, mn = 1

First we consider Left Hand Side (LHS),

= mn

= (sec θ + tan θ)( sec θ – tan θ)

We know that, a2 – b2 = (a + b) (a – b)

= sec2 θ – tan2 θ

Also we know that, 1 + tan2 θ = sec2 θ

= 1

Then, Right Hand Side = 1

Therefore, LHS = RHS

Hence it is proved that, mn = 1.

12. If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x2 – y2 = a2 – b2

Solution:-

From the question it is given that,

x = a sec θ + b tan θ

y = a tan θ + b sec θ

We have to prove that, x2 – y2 = a2 – b2

First we consider Left Hand Side (LHS),

= x2 – y2

= (a sec θ + b tan θ)2 – (a tan θ + b sec θ)2

We know that, (a + b)2 = a2 + 2ab + b2

= a2 sec2 θ + b2 tan2 θ + 2 ab sec θ tan θ – (a2 tan2 θ + b2 sec2 θ + 2ab sec θ tan θ)

Then,

= sec2 θ (a2 – b2) + tan2 θ (b2 – a2)

= (a2 – b2)[sec2 θ – tan2 θ]

Also we know that, sec2 θ – tan2 θ = 1

= a2 – b2

Then, Right Hand Side = a2 – b2

Therefore, LHS = RHS

Hence it is proved that, x2 – y2 = a2 – b2.

13. If tan A + sin A = m and tan A – sin A = n, prove that (m2 – n2)2 = 16mn

Solution:-

From the question it is given that,

tan A + sin A = m

tan A – sin A = n

We have to prove that, (m2 – n2)2 = 16mn

First we consider Left Hand Side (LHS),

= (m2 – n2)2

= [(tan A + sin A)2 – (tan A – sin A)2]2

=[{(tan A + sin A) – (tan A – sin A)}{(tan A + sin A) + (tan A – sin A)}]2

By simplification we get,

= [(2 sin A)(2 tan A)]2

= [4 sin A tan A]2

= 16sin2 A tan2 A

Then, Right Hand Side = 16mn

= 16(tan2 A – sin2 A)

= 16((sin2 A/cos2 A) – sin2 A)

= 16 sin2 A ((1 – cos2 A)/cos2 A)

= 16 sin2 A (sin2 A/cos2 A)

= 16 sin2 A tan2 A

Therefore, LHS = RHS

Hence it is proved that, (m2 – n2)2 = 16mn.

14. If sin A + cos A = √2, prove that sin A cos A = ½

Solution:-

From the question it is given that, sin A + cos A = √2

We have to prove that, sin A cos A = ½

We know that, (sin A + cos A)2 = sin2 A + cos2 A + 2 sin A cos A

So, 2 = 1 + 2 sin A cos A

2 sin A cos A = 1

sin A cos A = ½

15. If a sin2 θ + b cos2 θ = c and p sin2 θ + q cos2 θ = r, prove that (b – c)(r – p) = (c – a)(q – r).

Solution:-

From the question it is given that,

a sin2 θ + b cos2 θ = c

p sin2 θ + q cos2 θ = r

We have to prove that, (b – c)(r – p) = (c – a)(q – r)

Consider, LHS = (b – c)(r – p)

= (b – a sin2 θ + b cos2 θ)( p sin2 θ + q cos2 θ – p)

= [b(1 – cos2 θ) – a sin2 θ] [p(sin2 θ – 1) + q cos2 θ]

= [(b – a) sin2 θ][(q – p) cos2 θ]

= (b – a)(q – p)sin2 θ cos2 θ

Now consider, RHS = (c – a) (q – r)

= (a sin2 θ + b cos2 θ – a) (q – p sin2 θ – q cos2 θ)

= [(b – a) cos2 θ][(q – p) sin2 θ]

= (b – a)(q – p) sin2 θ cos2 θ

Therefore, LHS = RHS

Hence it is proved that, (b – c)(r – p) = (c – a)(q – r).

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*