# 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 BYJUâ€™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

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

1. Prove the following identities:

(i) (1 â€“ sin2 Î¸) sec2Î¸ = 1

Solution:-

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

= (1 â€“ sin2 Î¸) sec2Î¸

We know that, (1 â€“ sin2 Î¸) = cos2 Î¸

= cos2Î¸ sec2Î¸

Also we know that, sec2Î¸ = 1/cos2Î¸

= cos2 Î¸ Ã— (1/cos2Î¸)

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(ii) (1 â€“ cos2 Î¸) sec2Î¸ = tan2Î¸

Solution:-

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

= (1 â€“ cos2 Î¸) sec2Î¸

We know that, (1 â€“ cos2 Î¸) = sin2 Î¸

= sin2Î¸ sec2Î¸

Also we know that, sec2Î¸ = 1/cos2Î¸

= sin2 Î¸ Ã— (1/cos2Î¸)

= sin2Î¸/cos2Î¸

= tan2Î¸

Then, Right Hand Side (RHS) = tan2Î¸

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 Î¸(1 + tan Î¸) + cos Î¸ (1 + cot Î¸) = sec Î¸ + cosec Î¸

Solution:-

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

= sin Î¸(1 + tan Î¸) + cos Î¸ (1 + cot Î¸)

We know that, tan Î¸ = sin Î¸/cos Î¸, cot Î¸ = cos Î¸/sin Î¸

= sin Î¸(1 + (sin Î¸/cos Î¸)) + cos Î¸ (1 + (cos Î¸/sin Î¸))

= sin Î¸((cos Î¸ + sin Î¸)/cos Î¸)) + cos Î¸ ((sin Î¸ + cos Î¸)/sin Î¸))

= cos Î¸ + sin Î¸ ((sin Î¸/cos Î¸) + (cos Î¸/sin Î¸))

= cos Î¸ + sin Î¸ ((1/sin Î¸)(1/cos Î¸))

= sec Î¸ + cosec Î¸

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

Therefore, LHS = RHS

(v) (1 + cot Î¸ â€“ cosec Î¸) (1 + tan Î¸ + sec Î¸) = 2

Solution:-

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

= (1 + cot Î¸ â€“ cosec Î¸) (1 + tan Î¸ + sec Î¸)

We know that,

cot Î¸ = sin Î¸/cos Î¸, cosec Î¸ = 1/cos Î¸, tan Î¸ = cos Î¸/sin Î¸, sec Î¸ = 1/sin Î¸

= (1 + (sin Î¸/cos Î¸) + (1/cos Î¸)) (1 + (cos Î¸/sin Î¸) â€“ (1/sin Î¸))

Taking LCM we get,

= ((cos Î¸ + sin Î¸ + 1)/cos Î¸) ((sin Î¸ + cos Î¸ – 1)/sin Î¸)

= ((sin Î¸ + cos Î¸)2 â€“ 12)/(sin Î¸ cos Î¸)

= (1 + 2 sin Î¸ cos Î¸ – 1)/(sin Î¸ cos Î¸)

= (2 sin Î¸ cos Î¸)/(sin Î¸ cos Î¸)

By simplification we get,

= 2

Then, Right Hand Side (RHS) = 2

Therefore, LHS = RHS

(vi) sin Î¸ cot Î¸ + sin Î¸ cosec Î¸ = 1 + cos Î¸

Solution:-

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

= sin Î¸ cot Î¸ + sin Î¸ cosec Î¸

We know that, cot Î¸ = cos Î¸/sin Î¸, cosec Î¸ = 1/sin Î¸

= sin Î¸ (cos Î¸/sin Î¸) + sin Î¸ (1/sin Î¸)

= cos Î¸ + 1

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

Therefore, LHS = RHS

(vii) sec A (1 â€“ sin A) (sec A + tan A) = 1

Solution:-

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

= sec A (1 â€“ sin A) (sec A + tan A)

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

= (1/cos A) (1 â€“ sin A) ((1/cos A) + (sin A/cos A))

= ((1 â€“ sin A)/cos A) ((1 + sin A)/cos A)

By simplification we get,

= (1 â€“ 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 â€“ sin2A)/cos2 A

= cos2 A/cos2 A

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(ix) cosec Î¸ (1 + cos Î¸) (cosec Î¸ â€“ cot Î¸) = 1

Solution:-

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

= cosec Î¸ (1 + cos Î¸) (cosec Î¸ â€“ cot Î¸)

We know that, cosec = 1/sin Î¸, cot Î¸ = cos Î¸/sin Î¸

= (1/sin Î¸) (1 + cos Î¸) ((1/sin Î¸) â€“ (cos Î¸/sin Î¸))

= ((1 + cos Î¸)/sin Î¸) ((1 â€“ cos Î¸)/sin Î¸)

= ((1 â€“ cos2 Î¸)/sin2 Î¸)

= sin2 Î¸/sin2 Î¸

= 1

Then, Right Hand Side (RHS) = 1

Therefore, LHS = RHS

(x) (sec A â€“ 1)/(sec A + 1) = (1 â€“ cos A)/(1 + cos A)

Solution:-

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

= (sec A â€“ 1)/(sec A + 1)

We know that, sec A = 1/cos A

= ((1/cos A) â€“ 1)/((1/cos A) + 1)

By simplification we get,

= (1 â€“ cos A)/(1 + cos A)

Then, Right Hand Side (RHS) = (1 â€“ cos A)/(1 + cos A)

Therefore, LHS = RHS

(xi) (1 + sin A)/(1 â€“ sin A) = (cosec A + 1)/(cosec A – 1)

Solution:-

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

= (1 + sin A)/(1 â€“ 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 â€“ sin A)

Therefore, LHS = RHS

(xii) cos A/(1 + sin A) = sec A â€“ tan A

Solution:-

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

= sec A â€“ tan A

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

= (1/cos A) â€“ (sin A/cos A)

= (1 â€“ sin A)/cos A

= ((1 â€“ sin A)/cos A) ((1 + sin A)/(1 + sin A))

By cross multiplication we get,

= (1 â€“ 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 Î¸ + sec Î¸ – 1)/(tan Î¸ â€“ sec + 1) = (1 + sin Î¸)/cos Î¸

Solution:-

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

= (tan Î¸ + sec Î¸ – 1)/(tan Î¸ â€“ sec + 1)

The above terms can be written as,

= (tan Î¸ + sec Î¸ â€“ (sec2 Î¸ â€“ tan2 Î¸))/(1 + tan Î¸ â€“ sec Î¸)

= [tan Î¸ + sec Î¸ â€“ {(sec Î¸ + tan Î¸) (sec Î¸ â€“ tan Î¸)}]/(1 + tan Î¸ â€“ sec Î¸)

= [(tan Î¸ + sec Î¸) {1 â€“ (sec Î¸ â€“ tan Î¸)}]/(1 + tan Î¸ â€“ sec Î¸)

= [(tan Î¸ + sec Î¸) (1 + tan Î¸ â€“ sec Î¸)]/(1 + tan Î¸ â€“ sec Î¸)

= [tan Î¸ + sec Î¸]

= (1 + sin Î¸)/ cos Î¸

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

Therefore, LHS = RHS

2. Prove the following identities:

(i) sin2 A cos2 B â€“ cos2 A sin2 B = sin2 A â€“ sin2 B

Solution:-

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

= sin2 A cos2 B â€“ cos2 A sin2 B

We know that, cos2 B = (1 â€“ sin2 B),

= sin2 A (1 â€“ sin2 B) â€“ (1 â€“ sin2 A) sin2 B

= sin2 A â€“ sin2 A sin2 B â€“ sin2 B + sin2 A sin2 B

On simplification we get,

= sin2 A â€“ sin2 B

Then, Right Hand Side (RHS) = sin2 A â€“ sin2 B

Therefore, LHS = RHS

(ii). (1 â€“ tan A)2 + (1 + tan A)2 = 2 sec2 A

Solution:-

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

= (1 â€“ tan A)2 + (1 + tan A)2

Expanding the above terms we get,

= 1 + tan2 A â€“ 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 â€“ cosec2 A = cot4 A + cot4 A

Solution:-

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

= cosec4 A â€“ 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 â€“ 1, cot2 A + 1 = cosec2 A

So, (cosec2 A â€“ 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 â€“ sin4 A = 2 cos2 A â€“ 1

Solution:-

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

= cos4 A â€“ sin4 A

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

= (cos2 A â€“ sin2 A)(cos2 A + sin2 A)

= (cos2 A â€“ (1 â€“ cos2 A))

= 2 cos2 A â€“ 1

Then, Right Hand Side (RHS) = 2 cos2 A â€“ 1

Therefore, LHS = RHS

(vii) (sec A â€“ cos A) (sec A + cos A) = sin2 A + tan2 A

Solution:-

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

= (sec A â€“ cos A) (sec A + cos A)

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

= (sec2 A â€“ cos2 A)

Also we know that, sec2 A = 1 + tan2 A, cos2 A = 1 â€“ sin2 A

= 1 + tan2 A â€“ (1 â€“ 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 â€“ sin A)2 = 2

Solution:-

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

= (cos A + sin A)2 + (cos A â€“ sin A)2

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

= cos2 A + sin2 A + 2 cos A sin A + cos2 A + sin2 A â€“ 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 â€“ sin A) (sec A â€“ cos A) (tan A + cot A) = 1

Solution:-

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

= (cosec A â€“ sin A) (sec A â€“ 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) â€“ cos A) (tan A + (1/tan A))

= ((1 â€“ sin2 A)/sin A) ((1 â€“ 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, cosec2Â 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 â€“ tan A)) + (sin A/(1 â€“ cot A)) = sin A + cos A

Solution:-

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

= (cos A/(1 â€“ tan A)) + (sin A/(1 â€“ cot A))

= (cos A/(1 â€“ (sin A/cos A))) + (sin A/(1 â€“ (cos A/sin A)))

= (cos A/((cos A â€“ sin A)/cos A)) + (sin A/((sin A â€“ cos A)/sin A))

= (cos2A/(cos A â€“ sin A)) + (sin2 A/(sin A â€“ cos A))

= (cos2 A â€“ sin2 A)/(cos A â€“ sin A)

= sin A + cos A

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

Therefore, LHS = RHS

(xiii) (cosec A â€“ sin A) (sec A â€“ cos A) = 1/(tan A + cot A)

Solution:-

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

= (cosec A â€“ sin A) (sec A â€“ cos A)

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

= ((1/sin A) â€“ sin A) ((1/cos) â€“ cos A)

= ((1 â€“ sin2 A)/sin A)((1 â€“ 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 â€“ 2 sin2 A cos2 A

Solution:-

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

= sin4 A + cos4 A

= 1 â€“ 2 sin2 A cos2 A

= sin4 A + cos4Â A + 2 sin2 A cos2Â A

= (sin2A)2+(cos2A)2+ 2 sin2A cos2A âˆ’ 2sin2Acos2AÂ  Â

[Adding and subtracting 2Â sin2AÂ cos2A]

= (sin2A + cos2A)2 â€“ 2 sin2 A cos2 A

= 1 â€“ 2 sin2 A cos2 A

Then, Right Hand Side (RHS) = 1 â€“ 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 â€“ cos A) = (cosec A + cot A)2

Solution:-

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

= (1 + cos A)/(1 â€“ cos A)

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

= ((1 + cos A)/(1 â€“ cos A)) ((1 + cos A)/(1 + cos A))

By cross multiplication we get,

= (1 + cos A)2/(1 â€“ 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 â€“ cot A = (1 â€“ 2 cos2 A)/(sin A cos A)

Solution:-

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

= tan A â€“ cot A

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

Then,

= (sin A/cos A) â€“ (cos A /sin A)

= (sin2 A â€“ cos2 A)/(sin A cos A)

We know that, sin2 A = 1 â€“ cos2 A

= (1 â€“ cos2 A â€“ cos2 A)/(sin A cos A)

So,

= (1 â€“ 2 cos2 A)/(sin A cos A)

Then, Right Hand Side (RHS) = (1 â€“ 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 â€“ cot A))

Solution:-

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

= cosec A + cot A

Now multiply and divide by cosec A â€“ cot A we get,

= ((cosec A + cot A)/1) ((cosec A â€“ cot A)/(cosec A â€“ cot A))

By cross multiplication we get,

= (cosec2 A â€“ cot2 A)/(cosec A â€“ cot A)

= (1 + cot2 A â€“ cot2 A)/(cosec A â€“ cot A)

= (1/(cosec A â€“ cot A))

Then, Right Hand Side (RHS) = (1/(cosec A â€“ 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 â€“ cosec A)/(cosec2 A – 1)

We know that, cosec2 A â€“ 1 = cot2 A

= 2 cosec2 A/cot2 A

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

= (2/sin2 A)/(cos2Â 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 â€“ cos A) = (tan2 A)/(sec A â€“ 1)2

Solution:-

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

= (1 + cos A)/(1 â€“ cos A)

We know that, cos A = 1/sec A

Then,

= (1 + (1/sec A))/(1 â€“ (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 â€“ cosec A)2 = ((1 â€“ cos A)/(1 + cos A))

Solution:-

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

= (cot A â€“ cosec A)2

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

= ((cos A/sin A) â€“ (1/sin A))2

= ((cos A – 1)/sin A)2

= (cos A – 1)2/sin2 A

= (cos A – 1)2/1 â€“ cos2 A

= (-(1 â€“ cos A)2/((1 â€“ cos A) (1 + cos A))

= ((1 â€“ cos A) (1 â€“ cos A))/((1 â€“ cos A)(1 + cos A))

= (1 â€“ cos A)/(1 + cos A)

Then, Right Hand Side (RHS) = (1 â€“ cos A)/(1 + cos A)

Therefore, LHS = RHS

4. Prove the following identities:

(i) âˆš(cosec2 q – 1) = cos q cosec q

Solution:-

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

= âˆš(cosec2 q – 1)

We know that, cosec2 q â€“ 1 = cot2 q

= âˆš(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) âˆš((1 + sin q)/(1 â€“ sin q)) + âˆš((1 â€“ sin q)/(1 + sin q)) = 2 sec q

Solution:-

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

= âˆš((1 + sin q)/(1 â€“ sin q)) + âˆš((1 â€“ sin q)/(1 + sin q))

Then,

= âˆš(((1 + sin q)/(1 â€“ sin q)) ((1 + sin q)/(1 + sin q))) + âˆš(((1 â€“ sin q)/(1 + sin q)) ((1 + sin q)/(1 + sin q)))

= âˆš((1 + sin q)2/(1 â€“ sin2 q)) + âˆš((1 â€“ sin q)2/(1 â€“ sin2 q))

We know that, 1 â€“ sin2 q = cos2 q

= âˆš((1 + sin q)2/cos2 q) + âˆš((1 â€“ sin q)2/cos2 q)

= ((1 + sin q)/cos q) + ((1 â€“ sin q)/cos q)

= 2/cos q

= 2 sec q

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

Therefore, LHS = RHS

(iii) âˆš((1 â€“ cos A)/(1 + cos A)) = (sin A/(1 + cos A))

Solution:-

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

= âˆš((1 â€“ cos A)/(1 + cos A))

Then,

= âˆš(((1 â€“ cos A)/(1 + cos A)) ((1 + cos A)/(1 + cos A)))

= âˆš((1 â€“ cos2 A)/(1 + cos A)2)

= âˆš((sin2Â A)/(1 + cos A)2)

= sin A/(1 + cos A)

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

Therefore, LHS = RHS

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

Solution:-

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

= âˆš((1 + cos A)/(1 – cos A))

Then,

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

= âˆš((1 + cos A)2/(1 â€“ cos2 A))

We know that, 1 â€“ cos2Â A = sin2 A

= âˆš((1 + cos2 A)/(sin2 A))

= âˆš((1 + cos A)/(sin A))2

= âˆš((1/sin A) + (cos A/sin A))2

= âˆš(cosec A + cot A)2

= cosec A + cot A

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

Therefore, LHS = RHS

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

Solution:-

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

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

Then,

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

By simplification we get,

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

We know that, sec2Â q â€“ 1 = tan2Â q

= âˆš((sec q – 1)2/tan2Â q) + âˆš((sec q + 1)2/tan2Â q)

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

= (sec q â€“ 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 Î¸ â€“ tan Î¸)2 = (1 â€“ sin Î¸)/(1 + sin Î¸)

Solution:-

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

= (sec Î¸ â€“ tan Î¸)2

We know that, sec Î¸ = 1/cos Î¸, tan Î¸ = sin Î¸/cos Î¸

= ((1/cos Î¸) â€“ (sin Î¸/cos Î¸))2

By taking LCM we get,

= ((1 â€“ sin Î¸)/cos Î¸)2

= (1 – sin Î¸)2/cos2 Î¸

Also we know that, cos2 Î¸ = 1 â€“ sin2 Î¸

= (1 â€“ sin Î¸)2/(1 â€“ sin2 Î¸)

= (1 â€“ sin Î¸)2/((1 â€“ sin Î¸)(1 + sin Î¸))

= (1 â€“ sin Î¸)/(1 + sin Î¸)

Then, Right Hand Side (1 â€“ sin Î¸)/(1 + sin Î¸)

Therefore, LHS = RHS

(ii) (1/(sin A + cos A)) + (1/(sin A â€“ cos A)) = 2 sin A/(1 â€“ 2 cos2 A)

Solution:-

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

= (1/(sin A + cos A)) + (1/(sin A â€“ cos A))

By taking LCM we get,

= (sin A â€“ cos A + sin A + cos A)/(sin2 A â€“ cos2 A)

= 2 sin A/(1 â€“ cos2 A â€“ cos2 A)

= 2 sin A/(1 â€“ 2 cos2 A)

Then, Right Hand Side 2 sin A/(1 â€“ 2 cos2 A)

Therefore, LHS = RHS

(iii) ((sin A + cos A)/(sin A – cos A)) + ((sin A â€“ 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 â€“ cos A)/(sin A + cos A))

By taking LCM we get,

= ((sin A + cos A)2 + (sin A â€“ cos A)2)/((sin A + cos A)(sin A â€“ cos A))

Then,

=(sin2Â A + cos2 A + 2 sin A cos A + sin2 A + cos2 A â€“ 2 sin cos A)/(sin2 A â€“ cos2 A)

= (2 sin2 A + 2 cos2 A + 2 sin A cos A â€“ 2 sin cos A)/(sin2 A â€“ cos2 A)

= (2 sin2 A + 2 cos2 A)/(sin2 A â€“ cos2 A)

= 2(sin2 A + cos2 A)/(sin2 A â€“ cos2 A)

We know that, sin2 A + cos2 A = 1

= 2(1)/(sin2 A â€“ cos2 A)

= 2/(sin2 A â€“ cos2 A)

= 2/(sin2 A â€“ (1 – sin2 A))

= 2/(2 sin2 A – 1)

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

Therefore, LHS = RHS

(iv) tan2 A â€“ tan2 B = (sin2A â€“ sin2 B)/(cos2 A cos2 B)

Solution:-

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

= tan2 A â€“ tan2 B

We know that, tan2 Î¸ = sin2 Î¸ /cos2 Î¸,

= (sin2 A cos2 B â€“ cos2 A sin2 B)/(cos2 A cos2 B)

Then,

= (((1 â€“ cos2 A) cos2 B) â€“ (cos2 A sin2 B))/(cos2 A cos2 B)

= (cos2 B â€“ cos2 A cos2 B â€“ cos2 A + cos2 A cos2 B)/(cos2 A cos2 B)

By simplification we get,

= (cos2 B â€“ cos2 A)/(cos2 A cos2 B)

= ((1 â€“ sin2 B) â€“ (1 â€“ sin2 A))/(cos2 A cos2 B)

= (sin2 A â€“ sin2 B)/(cos2 A cos2 B)

Then, Right Hand Side (sin2 A â€“ sin2 B)/(cos2 A cos2 B)

Therefore, LHS = RHS

(v) (cos A/(1 â€“ tan A)) + (sin2 A/(sin A â€“ cos A)) = cos A + sin A

Solution:-

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

= (cos A/(1 â€“ tan A)) + (sin2 A/(sin A â€“ cos A))

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

= (cos A/(1 â€“ (sin A/cos A))) + (sin2 A/(sin A â€“ cos A))

Then,

= (cos A/((cos A â€“ sin A)/cos A)) + (sin2 A/(sin A â€“ cos A))

= (cos2 A/(cos A â€“ sin A)) â€“ (sin2 A/(cos A â€“ sin A))

= (cos2 A â€“ sin2 A)/(cos A â€“ sin A)

= ((cos A + sin A)(cos A â€“ sin A))/(cos A â€“ 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 â€“ 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 â€“ sin2 A)) + (1/sin2 A)

= (sin2 A + 1 â€“ sin2 A)/(sin2 A(1 â€“ sin2 A))

= 1/(sin2 A â€“ sin4 A)

Then, Right Hand Side = 1/(sin2 A â€“ sin4 A)

Therefore, LHS = RHS

(vii) ((cos3 A + sin3 A)/(cos A + sin A)) + ((cos3 A â€“ sin3 A)/(cos A â€“ sin A)) = 2

Solution:-

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

= ((cos3 A + sin3 A)/(cos A + sin A)) + ((cos3 A â€“ sin3 A)/(cos A â€“ sin A))

By taking LCM we get,

= ((cos3 A + sin3 A)(cos A â€“ sin A) + (cos3 A â€“ sin3 A)(cos A + sin A))/(cos2 A â€“ sin2 A)

= 2(cos4 A â€“ sin4 A)/(cos2 A â€“ sin2 A)

= (2(cos2 A + sin2 A)(cos2 A â€“ sin2 A))/(cos2 A â€“ 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 Î¸ + (1/cos Î¸))2 + (tan Î¸ â€“ (1/cos Î¸))2 = 2((1 + sin2 Î¸)/(1 â€“ sin2 Î¸))

Solution:-

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

= (tan Î¸ + (1/cos Î¸))2 + (tan Î¸ â€“ (1/cos Î¸))2

We know that, tan Î¸ = sin Î¸/cos Î¸

= ((sin Î¸/cos Î¸) + (1/cos Î¸))2 + ((sin Î¸/cos Î¸) â€“ (1/cos Î¸))2

= ((sin Î¸ + 1)/cos Î¸)2 + ((sin Î¸ – 1)/cos Î¸)2

= ((sin Î¸ + 1)2/cos2 Î¸) + ((sin Î¸ – 1)/cos2 Î¸)

= ((sin Î¸ + 1)2 + (sin Î¸ – 1)2)/cos2 Î¸

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

= (sin2 Î¸ + 1 + 2 sin Î¸ + sin2 Î¸ + 1 â€“ 2 sin Î¸)/(1 â€“ sin2 Î¸)

= 2(1 + sin2 Î¸)/(1 â€“ sin2 Î¸)

Then, Right Hand Side = 2(1 + sin2 Î¸)/(1 â€“ sin2 Î¸)

Therefore, LHS = RHS

(ix) ((sin A â€“ sin B)/(cos A + cos B)) + ((cos A â€“ cos B)/(sin A + sin B)) = 0

Solution:-

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

= ((sin A â€“ sin B)/(cos A + cos B)) + ((cos A â€“ cos B)/(sin A + sin B))

By taking LCM we get,

= (((sin A + sin B)(sin A â€“ sin B)) + ((cos A + cos B)(cos A â€“ cos B)))/((cos A + cos B)(sin A â€“ sin B))

By simplification we get,

= ((sin2 A â€“ sin2 B) + (cos2 A â€“ cos2 B))/((cos A + cos B)(sin A â€“ sin B))

= ((sin2 A + cos2 A) â€“ (sin2 B + cos2 B))/((cos A + cos B)(sin A â€“ sin B))

We know that, sin2 A + cos2 A = 1

= (1 – 1)/ ((cos A + cos B)(sin A â€“ sin B))

= 0/((cos A + cos B)(sin A â€“ 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 â€“ 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 â€“ cosec A + 1)

We know that, cosec2 A â€“ cot2 A = 1

= (cot A + cosec â€“ (cosec2 A â€“ cot2 A))/(cot A â€“ cosec A + 1)

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

= [cot A + cosec A â€“ ((cosec A â€“ cot A)(cosec A + cot A))]/(cot A â€“ cosec A + 1)

= (cot A + cosec A [1 â€“ cosec A + cot A])/(cot A â€“ 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 â€“ cos A)/(1 + cos A)

Then,

= ((1 â€“ cos A)/(1 + cos A)) Ã— ((1 + cos A)/(1 + cos A))

By simplification we get,

= (1 â€“ cos2 A)/(1 + cos A)2

Also we know that, 1 â€“ 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 â€“ cot A)2 = 2 cosec2 A

Solution:-

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

= (1 + cot A)2 + (1 â€“ cot A)2

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

= 1 + cot2 A + 2 cot A + 1 + cot2 A â€“ 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 Î¸/(tan Î¸ + cot Î¸)) = cos Î¸

Solution:-

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

= (cosec Î¸/(tan Î¸ + cot Î¸))

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

Then,

= (1/sin Î¸)/((sin Î¸/cos Î¸) + (cos Î¸/sin Î¸))

Taking LCM in the denominator we get,

= (1/sin Î¸)/((sin2 Î¸ + cos2 Î¸)/(cos Î¸ sin Î¸))

Also we know that, sin2 Î¸ + cos2 Î¸ = 1

= (1/sin Î¸)/(1/cos Î¸ sin Î¸)

= (1/sin Î¸) Ã— ((cos Î¸ sin Î¸)/1)

= cos Î¸

Then, Right Hand Side = cos Î¸

Therefore, LHS = RHS

(iii) (1 + tan2 Î¸) sin Î¸ cos Î¸ = tan Î¸

Solution:-

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

= (1 + tan2 Î¸) sin Î¸ cos Î¸

We know that, tan2 Î¸ = sin2 Î¸/cos2 Î¸

= (1 + (sin2 Î¸/cos2 Î¸)) sin Î¸ cos Î¸

Taking LCM we get,

= ((cos2 Î¸ + sin2 Î¸)/cos2 Î¸) sin Î¸ cos Î¸

Also we know that, sin2 Î¸ + cos2 Î¸ = 1

= (1/cos2 Î¸) sin Î¸ cos Î¸

= sin Î¸/cos Î¸

= tan Î¸

Then, Right Hand Side = tan Î¸

Therefore, LHS = RHS

(iv) ((1 + sin Î¸)/(cosec Î¸ â€“ cot Î¸)) â€“ ((1 â€“ sin Î¸)/(cosec Î¸ + cot Î¸)) = 2 (1 + cot Î¸)

Solution:-

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

= ((1 + sin Î¸)/(cosec Î¸ â€“ cot Î¸)) â€“ ((1 â€“ sin Î¸)/(cosec Î¸ + cot Î¸))

By taking LCM we get,

= [((1 + sin Î¸)(cosec Î¸ + cot Î¸)) â€“ ((1 â€“ sin Î¸)(cosec Î¸ â€“ cot Î¸))]/(cosec2 Î¸ â€“ cot2 Î¸)

We know that, 1 + cot2 Î¸ = cosec2 Î¸

= (cosec Î¸ + cot Î¸ + 1 + cos Î¸ â€“ cosec Î¸ + cot Î¸ + 1 â€“ cos Î¸)/(1 + cot2 Î¸ â€“ cot2 Î¸)

By simplification we get,

= 2 + 2 cot Î¸

Taking common terms outside we get,

= 2(1 + cot Î¸)

Then, Right Hand Side = 2(1 + cot Î¸)

Therefore, LHS = RHS

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

Solution:-

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

= (1 + cot A + tan A) (sin A â€“ 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 â€“ cos A)

By taking LCM we get,

= [(sin A cos A + cos2 A + sin2 A)/(sin A cos A)] (sin A â€“ cos A)

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

So,

= (sin3 A â€“ cos3 A)/(sin A cos A)

= (sin3 A/(sin A cos A)) â€“ (cos3 A/(sin A cos A))

= (sin2 A/cos A) â€“ (cos2 A/sin A)

= ((1/cos A) Ã— sin2 A) â€“ ((1/sin A) Ã— cos2 A)

= sec A sin2 A â€“ cosec A cos2 A

= (sec A/cosec2 A) â€“ (cosec A/sec2 A)

Then, Right Hand Side = (sec A/cosec2 A) â€“ (cosec A/sec2 A)

Therefore, LHS = RHS

(vi) 2(sin6 Î¸ + cos6 Î¸) â€“ 3 (sin4 Î¸ + cos4 Î¸) + 1 = 0

Solution:-

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

= 2(sin6 Î¸ + cos6 Î¸) â€“ 3 (sin4 Î¸ + cos4 Î¸) + 1

The above terms can be written as,

= 2[(sin2 Î¸)3 + (cos2 Î¸)3] â€“ 3 (sin4 Î¸ + cos4 Î¸) + 1

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

= 2[(sin2 Î¸ + cos2 Î¸) ((sin2 Î¸) + (cos2 Î¸)2 â€“ sin2 Î¸ cos2 Î¸)] â€“ 3(sin4 Î¸ + cos4 Î¸) + 1

= 2[(sin2 Î¸)2 + (cos2 Î¸)2 â€“ sin2 Î¸ cos2 Î¸] â€“ 3 (sin4 Î¸ + cos4 Î¸) + 1

= 2 sin4 Î¸ + 2 cos4 Î¸ â€“ 2 sin2 Î¸ cos2 Î¸ â€“ 3 sin4 Î¸ â€“ 3 cos4 Î¸ + 1

= – sin4 Î¸ â€“ cos4 Î¸ â€“ 2 sin2 Î¸ cos2 Î¸ + 1

= -(sin4 Î¸ + cos4 Î¸ + 2 sin2 Î¸ cos2 Î¸) + 1

= – (sin2 Î¸ + cos2 Î¸)2 + 1

Also we know that, sin2 Î¸ + cos2 Î¸ = 1

= – 1 + 1

= 0

Then, Right Hand Side = 0

Therefore, LHS = RHS

(vii) sin8 Î¸ â€“ cos8 Î¸ = (sin2 Î¸ â€“ cos2 Î¸)(1 â€“ 2 sin2 Î¸ cos2 Î¸)

Solution:-

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

= sin8 Î¸ â€“ cos8 Î¸

The above terms can be written as,

= (sin4 Î¸)2 â€“ (cos4 Î¸)2

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

= (sin4 Î¸ â€“ cos4 Î¸) (sin4 Î¸ + cos4 Î¸)

= ((sin2 Î¸)2 â€“ (cos2 Î¸)2) (sin4 Î¸ + cos4 Î¸)

= (sin2 Î¸ â€“ cos2 Î¸) (sin2 Î¸ + cos2 Î¸)(sin4 Î¸ + cos4 Î¸)

= (sin2 Î¸ â€“ cos2 Î¸)(sin4 Î¸ + cos4 Î¸)

= (sin2 Î¸ â€“ cos2 Î¸)((sin2 Î¸)2 + (cos2 Î¸)2 + 2sin2 Î¸ cos2 Î¸ â€“ 2sin2 Î¸ cos2 Î¸)

= (sin2 Î¸ â€“ cos2 Î¸)((sin2 Î¸ + cos2 Î¸)2 – 2 sin2 Î¸ cos2 Î¸)

= (sin2 Î¸ â€“ cos2 Î¸)(1 â€“ 2sin2 Î¸ cos2 Î¸)

Then, Right Hand Side = (sin2 Î¸ â€“ cos2 Î¸)(1 â€“ 2sin2 Î¸ cos2 Î¸)

Therefore, LHS = RHS

(viii) sec4 A â€“ sec2 A = (sin2 A/cos4 A)

Solution:-

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

= sec4 A â€“ sec2 A

We know that, sec A = 1/cos A

= (1/cos4 A) â€“ (1/cos2 A)

= (1 â€“ cos2 A)/cos4 A

Also we know that, 1 â€“ cos2 A = sin2 A

= sin2 A/cos4 A

Then, Right Hand Side = sin2 A/cos4 A

Therefore, LHS = RHS

(ix) (tan2 Î¸/(tan2 Î¸ – 1)) + (cosec2 Î¸/(sec2 Î¸ â€“ cosec2 Î¸)) = (1/(sin2 Î¸ â€“ cos2 Î¸))

Solution:-

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

= (tan2 Î¸/(tan2 Î¸ – 1)) + (cosec2 Î¸/(sec2 Î¸ â€“ cosec2 Î¸))

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

= ((sin2 Î¸/cos2 Î¸)/((sin2 Î¸/cos2 Î¸) – 1)) + ((1/sin2 Î¸)/((1/cos2 Î¸) â€“ (1/sin2 Î¸)))

= (sin2 Î¸/(sin2 Î¸ â€“ cos2 Î¸)) + ((1/sin2 Î¸)/((sin2 Î¸ â€“ cos2 Î¸)/(cos2 Î¸ sin2 Î¸)))

= (sin2 Î¸/(sin2 Î¸ â€“ cos2 Î¸)) + (cos2 Î¸/(sin2 Î¸ â€“ cos2 Î¸))

= (sin2 Î¸ + cos2 Î¸)/(sin2 Î¸ â€“ cos2 Î¸)

= 1/(sin2 Î¸ â€“ cos2 Î¸)

Then, Right Hand Side = 1/(sin2 Î¸ â€“ cos2 Î¸)

Therefore, LHS = RHS

(x) (sec2 Î¸ â€“ sin2 Î¸)/(tan2 Î¸) = cosec2 Î¸ â€“ cos2 Î¸

Solution:-

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

= (sec2 Î¸ â€“ sin2 Î¸)/(tan2 Î¸)

We know that, sec2 Î¸ = 1/cos2 Î¸, tan2 Î¸ = sin2 Î¸/cos2 Î¸

= ((1/cos2 Î¸) â€“ sin2 Î¸)/(sin2 Î¸/cos2 Î¸)

= ((1 â€“ sin2 Î¸ cos2 Î¸)/cos2 Î¸)/(sin2 Î¸/cos2 Î¸)

= (1 â€“ sin2 Î¸ cos2 Î¸)/sin2 Î¸

= (1/sin2 Î¸) â€“ ((sin2 Î¸ cos2 Î¸)/sin2 Î¸)

= cosec2 Î¸ â€“ cos2 Î¸

Then, Right Hand Side = cosec2 Î¸ â€“ cos2 Î¸

Therefore, LHS = RHS

(xi) ((cos3 Î¸ + sin3 Î¸)/(cos Î¸ + sin Î¸)) + ((cos3 Î¸ â€“ sin3 Î¸)/(cos Î¸ â€“ sin Î¸)) = 2

Solution:-

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

= ((cos3 Î¸ + sin3 Î¸)/(cos Î¸ + sin Î¸)) + ((cos3 Î¸ â€“ sin3 Î¸)/(cos Î¸ â€“ sin Î¸))

By taking LCM we get,

= [((cos3 Î¸ + sin3 Î¸)(cos Î¸ â€“ sin Î¸)) + ((cos3 Î¸ â€“ sin3 Î¸)(cos Î¸ + sin Î¸))]/((cos Î¸ + sin Î¸)(cos Î¸ â€“ sin Î¸))

Then,

= (cos4 Î¸ â€“ cos3 Î¸ sin Î¸ + sin3 Î¸ cos Î¸ â€“ sin4 Î¸ + cos4 Î¸ + cos3 Î¸ sin Î¸ â€“ sin3 Î¸ cos Î¸ â€“ sin4 Î¸)/(cos2 Î¸ â€“ sin2 Î¸)

By simplification we get,

= (2 cos4 Î¸ â€“ 2 sin4 Î¸)/(cos2 Î¸ â€“ sin2 Î¸)

Taking common outside,

= 2(cos4 Î¸ â€“ sin4 Î¸)/(cos2 Î¸ â€“ sin2 Î¸)

= (2(cos2 Î¸ + sin2 Î¸)(cos2 Î¸ â€“ sin2 Î¸))/(cos2 Î¸ â€“ sin2 Î¸)

= 2(cos2 Î¸ + sin2 Î¸)

We know that, sin2 Î¸ + cos2 Î¸ = 1

= 2

Then, Right Hand Side = 2

Therefore, LHS = RHS

(xii) (tan Î¸ + sin Î¸)/(tan Î¸ â€“ sin Î¸) = (sec Î¸ + 1)/(sec Î¸ – 1)

Solution:-

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

= (tan Î¸ + sin Î¸)/(tan Î¸ â€“ sin Î¸)

We know that, tan Î¸ = sin Î¸/cos Î¸

= ((sin Î¸/cos Î¸) + sin Î¸)/((sin Î¸/cos Î¸) â€“ sin Î¸)

= (sin Î¸ + sin Î¸ cos Î¸)/(sin Î¸ â€“ sin Î¸ cos Î¸)

= (sin Î¸(1 + cos Î¸))/(sin Î¸(1 â€“ cos Î¸))

= (1 + cos Î¸)/(1 â€“ cos Î¸)

Also we know that, cos Î¸ = 1/sec Î¸

= (1 + (1/cos Î¸))/(1 â€“ (1/sec Î¸))

= ((sec Î¸ + 1)/sec Î¸)/((sec Î¸ – 1)sec Î¸)

= (sec Î¸ + 1)/(sec Î¸ – 1)

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

Therefore, LHS = RHS

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

Then,

m2 â€“ n2 = (a sec A + b tan A)2 â€“ (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 â€“ (a2 tan2 A + b2 sec2 A + 2ab sec A tan A)

= sec2 A(a2 â€“ b2) + tan2 A(b2 â€“ a2)

= (a2 â€“ b2)[sec2 A â€“ tan2 A]

Also we know that, sec2 A â€“ tan2 A = 1

= (a2 â€“ b2)

Hence it is proved that, m2 â€“ n2 = a2 â€“ 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 Î¸, y = b cot Î¸, prove that (a2/x2) â€“ (b2/y2) = 1.

Solution:-

From the question it is given that,

x = a cos Î¸

y = b cot Î¸

We have to prove that, (a2/x2) â€“ (b2/y2) = 1

First we consider Left Hand Side (LHS),

= (a2/x2) â€“ (b2/y2)

= (a2/a2cos2 Î¸) â€“ (b2/b2 cot2 Î¸)

= (1/cos2 Î¸) â€“ (1/cot2 Î¸)

= sec2 Î¸ â€“ tan2 Î¸

We know that, 1 + tan2 Î¸ = sec2 Î¸

= 1

Then, Right Hand Side = 1

Therefore, LHS = RHS

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

11. If sec Î¸ + tan Î¸ = m, sec Î¸ â€“ tan Î¸ = n, prove that mn = 1.

Solution:-

From the question it is given that,

sec Î¸ + tan Î¸ = m

sec Î¸ â€“ tan Î¸ = n

We have to prove that, mn = 1

First we consider Left Hand Side (LHS),

= mn

= (sec Î¸ + tan Î¸)( sec Î¸ – tan Î¸)

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

= sec2 Î¸ â€“ tan2 Î¸

Also we know that, 1 + tan2 Î¸ = sec2Â Î¸

= 1

Then, Right Hand Side = 1

Therefore, LHS = RHS

Hence it is proved that, mn = 1.

12. If x = a sec Î¸ + b tan Î¸ and y = a tan Î¸ + b sec Î¸, prove that x2 â€“ y2 = a2 â€“ b2

Solution:-

From the question it is given that,

x = a sec Î¸ + b tan Î¸

y = a tan Î¸ + b sec Î¸

We have to prove that, x2 â€“ y2 = a2 â€“ b2

First we consider Left Hand Side (LHS),

= x2 â€“ y2

= (a sec Î¸ + b tan Î¸)2 â€“ (a tan Î¸ + b sec Î¸)2

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

= a2 sec2 Î¸ + b2 tan2 Î¸ + 2 ab sec Î¸ tan Î¸ â€“ (a2 tan2 Î¸ + b2 sec2 Î¸ + 2ab sec Î¸ tan Î¸)

Then,

= sec2 Î¸ (a2 â€“ b2) + tan2 Î¸ (b2 â€“ a2)

= (a2 â€“ b2)[sec2 Î¸ â€“ tan2 Î¸]

Also we know that, sec2 Î¸ â€“ tan2 Î¸ = 1

= a2 â€“ b2

Then, Right Hand Side = a2 â€“ b2

Therefore, LHS = RHS

Hence it is proved that, x2 â€“ y2 = a2 â€“ b2.

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

Solution:-

From the question it is given that,

tan A + sin A = m

tan A â€“ sin A = n

We have to prove that, (m2 â€“ n2)2 = 16mn

First we consider Left Hand Side (LHS),

= (m2 â€“ n2)2

= [(tan A + sin A)2 â€“ (tan A – sin A)2]2

=[{(tan A + sin A) â€“ (tan A â€“ sin A)}{(tan A + sin A) + (tan A â€“ 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 â€“ sin2 A)

= 16((sin2 A/cos2 A) â€“ sin2 A)

= 16 sin2 A ((1 â€“ 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 â€“ n2)2 = 16mn.

14. If sin A + cos A = âˆš2, prove that sin A cos A = Â½

Solution:-

From the question it is given that, sin A + cos A = âˆš2

We have to prove that, sin A cos A = Â½

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 = Â½

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

Solution:-

From the question it is given that,

a sin2 Î¸ + b cos2 Î¸ = c

p sin2 Î¸ + q cos2 Î¸ = r

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

Consider, LHS = (b – c)(r – p)

= (b – a sin2 Î¸ + b cos2 Î¸)( p sin2 Î¸ + q cos2 Î¸ – p)

= [b(1 â€“ cos2 Î¸) â€“ a sin2 Î¸] [p(sin2 Î¸ – 1) + q cos2 Î¸]

= [(b – a) sin2 Î¸][(q – p) cos2 Î¸]

= (b – a)(q – p)sin2 Î¸ cos2 Î¸

Now consider, RHS = (c – a) (q â€“ r)

= (a sin2 Î¸ + b cos2 Î¸ – a) (q â€“ p sin2 Î¸ â€“ q cos2 Î¸)

= [(b – a) cos2 Î¸][(q – p) sin2 Î¸]

= (b – a)(q – p) sin2 Î¸ cos2 Î¸

Therefore, LHS = RHS

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