### Chapter 2 Inverse Trigonometric Functions – Exercise 2.2 | Set 1

### Find the values of each of the following:

### Question 11. tan^{−1}[2cos(2sin^{−1}1/2)]

**Solution:**

Let us assume that sin

^{−1}1/2 = xSo, sinx = 1/2

Therefore, x =

π/6 = sin^{−1}1/2Therefore,

tan^{−1}[2cos(2sin^{−1}1/2)] =tan^{−1}[2cos(2 *π/6)]=

tan^{−1}[2cos(π/3)]Also,

cos(π/3) = 1/2Therefore,

tan^{−1}[2cos(π/3)] =tan^{−1}[(2 * 1/2)]=

tan^{−1}[1] =π/4

### Question 12. cot(tan^{−1}a + cot^{−1}a)

**Solution:**

We know, tan

^{−1}x + cot^{−1}x =π/2Therefore, cot(tan

^{−1}a + cot^{−1}a) = cot(π/2) =0

### Question 13.

**Solution:**

We know, 2tan

^{-1}x = and 2tan^{-1}y =

=

tan(1/2)[2(tan^{−1}x+tan^{−1}y)]=

tan[tan^{−1}x+tan^{−1}y]Also,

tan^{−1}x+tan^{−1}y =Therefore, tan[

tan^{−1}x+tan^{−1}y] == (x + y)/(1 – xy)

### Question 14. If sin(sin^{−1}1/5 + cos^{−1}x) = 1 then find the value of x

**Solution:**

sin

^{−1}1/5 + cos^{−1}x = sin^{−1}1We know, sin

^{−1}1 = π/2Therefore, sin

^{−1}1/5 + cos^{−1}x = π/2sin

^{−1}1/5 = π/2 – cos^{−1}xSince, sin

^{−1}x + cos^{−1}x = π/2Therefore, π/2 – cos

^{−1}x = sin^{−1}xsin

^{−1}1/5 = sin^{−1}xSo, x = 1/5

### Question 15. If , then find the value of x

**Solution:**

We know, tan

^{−1}x + tan^{−1}y =2x

^{2 }– 4 = -32x

^{2 }– 4 + 3 = 02x

^{2 }– 1 = 0x

^{2 }= 1/2x = 1/√2, -1/√2

### Find the values of each of the expressions in Exercises 16 to 18.

### Question 16. sin ^{− 1}(sin2π/3)

**Solution:**

We know that sin

^{−1}(sinθ) =θ when θ ∈ [-π/2, π/2], butSo, sin

^{− 1}(sin2π/3) can be written assin

^{− 1}(sinπ/3) hereTherefore, sin

^{− 1}(sinπ/3) = π/3

### Question 17. tan^{−1}(tan3π/4)

**Solution:**

We know that

tanwhen but^{−1}(tanθ) = θSo, tan

^{−1}(tan3π/4) can be written as tan^{−1}(-tan(-3π/4))=

tan^{−1}[-tan(π – π/4)]=

tan^{−1}[-tan(π/4)]= –

tan^{−1}[tan(π/4)]= –

π/4where

### Question 18.

**Solution:**

Let us assume = x , so sinx = 3/5

We know,

cosx = 4/5

We know,

So,

tanx = 3/4

Also,

Hence,

tan

^{-1}x + tan^{-1}y =So,

= 17/6

### Question 19. cos^{−1}(cos7π/6) is equal to

### (i) 7π/6 (ii) 5π/6 (iii)π/3 (iv)π/6

**Solution:**

We know that cos

^{−1}(cosθ) = θ, θ ∈ [0, π]cos

^{−1}(cosθ) = θ, θ ∈ [0, π]Here, 7π/6 > π

So, cos

^{−1}(cos7π/6) can be written as cos^{−1}(cos(-7π/6))=

cos^{−1}[cos(2π –7π/6)] [cos(2π + θ) = θ]=

cos^{−1}[cos(5π/6)] where 5π/6 ∈ [0, π]Therefore,

cos^{−1}[cos(5π/6)] = 5π/6

### Question 20.

### (i) 1/2 (ii) 1/3 (iii) 1/4 (iv) 1

**Solution:**

Let us assume sin

^{-1}(-1/2)= x, so sinx = -1/2Therefore, x = -π/6

Therefore, sin[π/3 – (-π/6)]

= sin[π/3 + (π/6)]

= sin[3π/6]

= sin[π/2]

= 1

### Question 21. is equal to

### (i) π (ii) -π/2 (iii)0 (iv)2√3

**Solution:**

We know, cot(−x) = −cotx

Therefore, tan

^{-1}3 – cot^{-1}(-3) = tan^{-1}3 – [-cot^{-1}(3)]= tan

^{-1}3 + cot^{-1}3Since, tan

^{-1}x + cot^{-1}x = π/2Tan

^{-1}3 + cot^{-1}3 = -π/2

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