Frank Solutions for Class 10 Maths Chapter 21 Trigonometric Identities

Students can refer to Frank Solutions for Class 10 Maths Chapter 21 Trigonometric Identities, which help to gain conceptual knowledge. So, it is best to practise these solutions to enhance their understanding of the topics in the chapter. Solving the exercises in each chapter will ensure that students score good marks in the exams. Our solution module utilises various shortcut tips and practical examples to explain all the exercise questions in 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. Subject experts at BYJU’S have designed these solutions in a very lucid and clear manner to help students solve problems in the most efficient possible ways. Download the PDF of Frank Solutions for Class 10 Maths Chapter 21 from the link given below.

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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

= (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).