# Class 10 NCERT Solutions- Chapter 8 Introduction To Trigonometry – Exercise 8.4

• Last Updated : 13 Jan, 2021

### Question 1. Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A

Solution:

(i) sin A

We know that

cosec2A = 1 + cot2A

1/sin2A = 1 + cot2A

sin2A = 1/(1 + cot2A)

sin A = 1/(1+cot2A)1/2

(ii) sec A

sec2A = 1 + tan2A

Sec2A = 1 + 1/cot2A

sec2A = (cot2A + 1) / cot2A

sec A = (cot2A + 1)1/2 / cot A

(iii) tan A

tan A = 1 / cot A

tan A = cot -1 A

### Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A.

Solution:

(i) cos A

cos A = 1/sec A

(ii) sin A

We know that

sin2A = 1 – cos2A

Also , cos2A = 1 / sec2A

sin2A = 1 – 1 / sec2A

sin2A = (sec2A – 1) / sec2A

sin A = (sec2A – 1)1/2 / sec A

(iii) tan A

We know that

tan2A + 1 = sec2A

tan A = (sec2A – 1)½

(iv) cosec A

We know

cosec A = 1/ sinA

cosec A = sec A / (sec2A – 1)½

(v) cot A

We know

cot A = cos A / sin A

cot A = (1/sec A) / ((sec2A – 1)1/2 / sec A)

cot A = 1 / (sec2A – 1)1/2

### Question 3. Evaluate:

(i) (sin2 63° + sin2 27°)/(cos2 17° + cos2 73°)

(ii) sin 25° cos 65° + cos 25° sin 65°

(i) ([sin(90-27)]2 + sin2 27) / ([cos(90-73)]2 + cos2 73)

We know that

sin(90-x) = cos x
cos(90-x) = sin x

(cos2(27) + sin2 27) / (sin2(73) + cos2 73)

Using

sin2A + cos2A = 1

1/1 = 1

(ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)]

Using

sin(90-x) = cos x
cos(90-x) = sin x

= [sin 25 * sin 25] + [cos 25 * cos 25]

= sin2 25 + cos2 25

= 1

### Question 4. Choose the correct option. Justify your choice.

Solution:

(i) 9 sec2 A – 9 tan2 A

(A) 1 (B) 9 (C) 8 (D) 0

Using sec2A – tan2A = 1

9 (sec2A – tan2A ) = 9(1)

Ans (B)

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

(A) 0 (B) 1 (C) 2 (D) –1

Simplifying all ratios

= (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ)

= ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ)

= ((cosθ + sinθ)2 – 1) / (sinθ cosθ)

= (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ)

= 2

Ans (C)

(iii) (sec A + tan A) * (1 – sin A)

(A) sec A (B) sin A (C) cosec A (D) cos A

Simplifying sec A and tan A

= (1/cos A + sin A/cos A)*(1 – sin A)

= ((1 + sin A)/cos A)*(1 – sin A)

= (1 – sin2A)/cos A

= cos2A / cos A

= cos A

Ans (D)

(iv) (1 + tan2A) / (1 + cot2A)

(A) sec2A (B) –1 (C) cot2A (D) tan2A

Simplifying tan A and cot A

= (1 + (sin2A / cos2A)) / (1 + (cos2A / sin2A))

= ((cos2A + sin2A) / cos2A) / ((cos2A + sin2A) / sin2A)

= sin2A / cos2A

= tan2A

Ans (D)

### Question 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution:

(i) (cosec θ – cot θ)2 = (1 – cosθ) / (1 + cosθ)

Solving LHS

Simplifying cosec θ and cot θ

= (1-cos θ)2 / sin2θ

= (1-cos θ)2 / (1-cos2θ)

Using a2 – b2 = (a+b)*(a-b)

= (1-cos θ)2 / [(1-cos θ)*(1+cos θ)]

= (1-cos θ) / (1+cos θ) = RHS

Hence Proved

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

Solving LHS

Taking LCM

= (cos2A + (1+sin A)2) / ((1+sin A) cos A)

= (cos2A + 1 + sin2A + 2 sin A ) / ((1 + sin A)*cos A)

Using sin2A + cos2A = 1

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

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

= 2 / cos A

= 2 sec A = RHS

Hence Proved

(iii) (tan θ / (1 – cot θ)) + (cot θ / (1 – tan θ)) = 1 + sec θ*cosec θ

Solving LHS

Changing tan θ and cot θ in terms of sin θ and cos θ and simplifying

= ((sin2θ) / (cos θ *(sin θ-cos θ))) + ((cos2θ ) / (sin θ *(sin θ-cos θ)))

= (1 / (sin θ-cos θ)) * [(sin3θ – cos3θ) / (sin θ * cos θ)]

= (1 / (sin θ – cos θ)) * [ ((sin θ – cos θ) * ( sin2θ + cos2θ + sin θ * cos θ ))/(sin θ *cos θ)]

= (1+sin θ*cos θ) / (sin θ*cos θ)

= sec θ*cosec θ + 1 = RHS

Hence Proved

(iv) (1 + sec A) / sec A = sin2A / (1 – cos A)

Solving LHS

= cos A + 1

Solving RHS

= (1 – cos2A) / (1 – cos A)

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

= 1 + cos A = RHS

Hence Proved

(v) (cos A – sin A + 1) / (cos A + sin A – 1) = cosec A + cot A using the identity cosec2A = 1 + cot2A

Solving LHS

Multiplying numerator and denominator by (cot A – 1 + cosec A)

= (cot2A + 1 + cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – (1 + cosec2A – 2*cosec A))

= (2*cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – 1 – cosec2A + 2*cosec A)

= (2* cosec A *(cosec A + cot A) – 2*(cosec A + cot A)) / (cot2A – 1 – cosec2A + 2*cosec A)

= ((cosec A + cot A) * (2*cosec A – 2 )) / (2*cosec A – 2)

= cosec A + cot A = RHS

Hence Proved

(vi) [(1 + sin A) / (1 – sin A)]½ = sec A + tan A

Solving LHS

Multiplying numerator and denominator by (1+sinA)

= [((1 + sin A)*(1 + sin A)) / ((1 – sin A)*(1 + sin A))]½

= (1 + sin A) / (1 – sin2A)½

= (1 + sin A) / (cos2A)1/2

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

= sec A + tan A = RHS

Hence Proved

(vii) (sin θ – 2 sin3θ) / (2 cos3θ – cos θ) = tan θ

Solving LHS

= (sin θ * (1 – 2*sin2θ)) / (cos θ * (2*cos2θ – 1))

= (sin θ * (1 – 2*sin2θ )) / (cos θ * (2*(1 – sin2θ) – 1))

= (sin θ *(1 – 2*sin2θ)) / (cos θ * (1 – 2*sin2θ))

= tan θ = RHS

Hence Proved

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A

Solving LHS

= sin2A + cosec2A + 2*sin A *cosec A + cos2A + sec2A + 2*cos A *sec A

We know that cosec A = 1 / sin A

= 1 + 1 + cot2A + 1 + tan2A + 2 + 2

= 7 + tan2A + cot2A = RHS

Hence Proved

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

Solving LHS

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

= ((1 – sin2A) / sin A) * ((1 – cos2A) / cos A)

= (cos2A * sin2A) / (sin A * cos A)

= sin A * cos A

Solving RHS

Simplifying tan A and cot A

= (sin A * cos A) / ( sin2A + cos2A)

= sin A * cos A = RHS

Hence Proved

(x) (1 + tan2A) / (1 + cot2A ) = [(1 – tan A) / (1 – cot A)]2 = tan2A

Solving LHS

Changing cot A = 1 / tan A

= (tan2A * (1 + tan2A)) / (1 + tan2A) = tan2A = RHS

= [(1 – tan A) / (1 – cot A)]2 = (1 + tan2A – 2*tan A) / (1 + cot2A – 2*cot A)

= (sec2A – 2*tan A) / (cosec2A – 2*cot A)

Solving this we get

= tan2A

Hence Proved

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