Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 | Set 2

Content of this article has been merged with Chapter 2 Inverse Trigonometric Functions - Exercise 2.2 as per the revised syllabus of NCERT.

Find the values of each of the following: 

Question 11. tan⁻¹[2cos(2sin⁻¹1/2​)]

Solution:

Let us assume that sin⁻¹1/2 = x

So, sinx = 1/2

Therefore, x = π​/6 = sin⁻¹1/2

Therefore, tan⁻¹[2cos(2sin⁻¹1/2​)] =  tan⁻¹[2cos(2 * π​/6)]

= tan⁻¹[2cos(π​/3)]

Also, cos(π/3​) = 1/2​

Therefore, tan⁻¹[2cos(π​/3)] = tan⁻¹[(2 * 1/2)]

= tan⁻¹[1] = π​/4 

Question 12. cot(tan⁻¹a + cot⁻¹a) 

Solution:

We know, tan⁻¹x + cot⁻¹x = π​/2

Therefore, cot(tan⁻¹a + cot⁻¹a) = cot(π​/2) =0

Question 13.  tan\frac{1}{2}[sin^{-1}\frac{2x}{1+x^2}+cos^{-1}\frac{1-y^2}{1+y^2}],|x|<1,y>0,xy<1

Solution:

We know, 2tan^-1x = sin^{-1}\frac{2 x}{1+x^2}    and 2tan^-1y =  cos^{-1}[\frac{1 - y^2 }{1+y^2}]

\therefore tan\frac{1}{2}[sin^{-1}\frac{2x}{1+x^2}+cos^{-1}\frac{1-y^2}{1+y^2}]    

= tan(1/2)​[2(tan⁻¹x + tan⁻¹y)]

= tan[tan⁻¹x + tan⁻¹y]

Also, tan⁻¹x + tan⁻¹y = tan^{-1}\frac{x+y}{1-xy}

Therefore, tan[tan⁻¹x + tan⁻¹y] = tan[tan^{-1}\frac{x+y}{1-xy}]

= (x + y)/(1 - xy)

Question 14. If sin(sin⁻¹1/5​ + cos⁻¹x) = 1 then find the value of x

Solution:

sin⁻¹1/5​ + cos⁻¹x = sin⁻¹1

We know, sin⁻¹1 = π/2

Therefore, sin⁻¹1/5​ + cos⁻¹x = π/2

sin⁻¹1/5​ = π/2 - cos⁻¹x

Since, sin⁻¹x​ + cos⁻¹x = π/2

Therefore, π/2 - cos⁻¹x = sin⁻¹x

sin⁻¹1/5​ = sin⁻¹x

So, x = 1/5

Question 15. If tan^{-1}\frac{x-1}{x-2} + tan^{-1}\frac{x+1}{x+2} = \frac{\pi}{4}    , then find the value of x

Solution:

We know, tan⁻¹x + tan⁻¹y = tan^{-1}\frac{x+y}{1-xy}

tan^{-1}\frac{x-1}{x-2} + tan^{-1}\frac{x+1}{x+2} =tan^{-1}\frac{\frac{x-1}{x-2}+\frac{x+1}{x+2}}{1-\frac{x-1}{x-2}.\frac{x+1}{x+2}} = \frac{\pi}{4}

tan^{-1}\frac{\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2)-(x+1)(x-1)}{(x-2)(x+2)}} = \frac{\pi}{4}

tan^{-1}\frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x+1)(x-1)} = \frac{\pi}{4}

tan^{-1}(\frac{x^2+2x-x-2+x^2-2x+x-2}{x^2-4-x^2+1}) = \frac{\pi}{4}

tan^{-1}(\frac{x^2+x-2+x^2-x-2}{-3}) = \frac{\pi}{4}

tan^{-1}(\frac{2x^2-4}{-3}) = \frac{\pi}{4}

\frac{2x^2-4}{-3} = tan(\frac{\pi}{4})

\frac{2x^2-4}{-3} = 1

2x² - 4 = -3

2x² - 4 + 3 = 0

2x² - 1 = 0

x² = 1/2

x = 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], but \frac{2 \pi}{3} > \frac{\pi}{2}

So, sin ^{− 1}(sin2π/3​) can be written as sin^{-1}[sin(\pi-\frac{2\pi}{3})]

 sin ^{− 1}(sinπ/3​)  here \frac{-\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}

Therefore, sin ^{− 1}(sinπ/3​) = π/3

Question 17. tan⁻¹(tan3π/4​)

Solution:

We know that tan⁻¹(tanθ) = θ when \theta \epsilon(\frac{-\pi}{2},\frac{\pi}{2})  but \frac{3 \pi}{4} > \frac{\pi}{2}

So, 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/4 where \frac{-\pi}{4} \epsilon(\frac{-\pi}{2},\frac{\pi}{2})

Question 18. tan(sin^{-1}\frac{3}{5} + cot^{-1}\frac{3}{2})

Solution:

Let us assume sin^{-1}\frac{3}{5}   = x , so sinx = 3/5 

We know, cosx = \sqrt{1-sin^2x}

\therefore cosx = \sqrt{1-(\frac{3}{5})^2}

cosx = \sqrt{1-\frac{9}{25}}

cosx = \sqrt{\frac{25-9}{25}}

cosx = \sqrt{\frac{16}{25}}

cosx = 4/5

We know, tanx = \frac{sinx}{cosx}

So, tanx = \frac{\frac{3}{5}}{\frac{4}{5}}

tanx = 3/4

Also, tan^{-1}\frac{1}{x} = cot^{-1}x

Hence, tan(sin^{-1}\frac{3}{5} + cot^{-1}\frac{3}{2}) = tan(tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3})

tan^-1x + tan^-1y = tan^{-1}\frac{x+y}{1-xy}

So, tan(tan^{-1}\frac{3}{4}+tan^{-1}\frac{2}{3}) = tan(tan^{-1}\frac {\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}.\frac{2}{3}})

= tan(tan^{-1}\frac{\frac{9+8}{12}}{\frac{12-6}{12}})

= tan(tan^{-1}\frac{17}{6})

= 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. sin[\frac{\pi}{3} - sin^{-1}(-\frac{1}{2} )]

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

Solution:

Let us assume sin^-1(-1/2)= x, so sinx = -1/2 

Therefore, x = -π/6​

Therefore, sin[π/3​ - (-π/6​)]

= sin[π/3​ + (π/6​)]

= sin[3π/6]

= sin[π/2]

= 1

Question 21. tan^{-1}\sqrt{3} - cot^{-1}(-\sqrt{3})   is equal to

(i) π    (ii) -π/2    (iii)0    (iv)2√3

Solution:

We know, cot(−x) = −cotx

Therefore, tan^-13 - cot^-1(-3) = tan^-13 - [-cot^-1(3)]

= tan^-13 + cot^-13

Since, tan^-1x + cot^-1x = π/2

Tan^-13 + cot^-13 = -π/2