Related Articles

# 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

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

My Personal Notes arrow_drop_up