Ex-6.2 Trigonometric Identities, Class 10, Maths RD Sharma Solutions | Extra Documents, Videos & Tests for Class 10 PDF Download

 Q1) If cosθ = 45, find all other trigonometric ratios of angle Θ.

Solution:

We have:

sinΘ =

Therefore, sinΘ = 3/5

Q2) If sinΘ = 1/√2, find all other trigonometric ratios of angle Θ.

Solution:

 We have,

=  cosΘ = 1/√2

=  cotΘ = 1/tanΘ = 1/1 = 1

Q3) If tanΘ = 1/√2, find the value of 

Solution:

We know that secΘ =

=  cosecΘ =

Substituting it in equation (1) we get

Q4) If tanΘ = 3/4, find the value of 

Solution:

We know that

secΘ = 

=  secΘ = 5/4

=  secΘ =

Therefore, Weget 

Q5) If tanΘ = 12/5, find the value of 

Solution:

cotΘ = 1/tanΘ = 

=  cosecΘ =

=  sinΘ = 1/cosecΘ =

i.e. We get 

Q6) If cotΘ = 1/√3, find the value of 

Solution:

cosecΘ =

=  cosecΘ = 2/√3

=  sinΘ = 1/cosecΘ =

=  and1/cotΘ = sinΘ/cosΘ = cosΘ = sinΘ × cotΘ = √3/2 × 1/√3 = 1/2

Therefore, on substituting we get

Q7) If cosecA = √2, find the value of  

Solution:

We know that cotA =

=  tanA = 1/cotA = 1/1 = 1

=  sinA = 1/cosecA = 1/√2

=  sinA = 1/√2

cosA =

On substituting we get:

Q8) If  cotΘ = √3, find the value of 

Solution:

cosecΘ =

sinΘ = 1/cosecΘ = 1/2cotΘ = cosΘ/sinΘ   .cosΘ = cotΘ.sinΘ

⇒ cosΘ = √3/2

=  secΘ = 1/cosΘ = 2/√3

On substituting we get:

=  21/8

Q9) If 3cosΘ = 1, find the value of 

Solution:

cosΘ = 1/3,sinΘ = 

tanΘ = sinΘ/cosΘ =

On substituting we get

=  40/4 = 10

Q10) If √3tanΘ = sinΘ, find the value of sin²Θ−cos²Θ.

Solution:

√3.sinΘ/cosΘ = sinΘ

=  cosΘ = √3/3⇒ 1/√3

=  sinΘ =

=  sin²Θ−cos²Θ = 

=  2/3−1/3 = 1/3

Q11) If cosecΘ = 13/12, find the value of 

Solution:

sinΘ = 1/cosecΘ =

=  cosΘ =

Q12) If sinΘ+cosΘ = √2cos(90^°−Θ), find cotΘ.

Solution:

=  sinΘ+cosΘ = √2sinΘ[cos(90−Θ) = sinΘ]

⇒ cosΘ = √2sinΘ − sinΘ ⇒ cosΘ = sinΘ(√2−1)

Divide both sides with sinΘ we get

=  cosΘ/sinΘ = sinΘ/sinΘ(√2−1)

=  cotΘ = √2−1.

Q-13. If 2sin²Θ – cos²Θ = 2, then find the value of Θ.

Solution.

2sin²Θ – cos²Θ = 2

⇒ 2sin²Θ−(1 – sin²Θ) = 2

⇒ 2sin²Θ – 1+sin²Θ = 2

⇒ 3sin²Θ = 3

⇒ sin²Θ = 1

⇒ sinΘ = 1

⇒ sinΘ = sin90°

⇒ Θ = 90°

Q-14. If √3tanΘ – 1 = 0, find the value of sin²Θ – cos²Θ.

Solution.

√3tanΘ – 1 = 0

⇒ √3tanΘ = 1

⇒ √3tanΘ = 1/√3

√3tanΘ = tan30°

Θ = 30°

Now,

sin²Θ – cos²Θ

=  sin²(30°) – cos²(30°)

=  (1/2)² – (√3/2)²

=  1/4 – 3/4  =  −2/4  =  −1/2