# Multiple choice (Inverse Trigonometry)

**choose the correct option in the following question:(class 12 inverse trigonometric functions multiple choice)**

**Question 1: The value of** **corresponding to principal branches is**

(a) -π/12 (b) 0

(c) π (d) π/3

**Answer (c)**

**Explanation:**

**Question 2: The value of** is

(a) (b)

(c) 1/x (d)

**Answer (d)**

**Explanation:**

Let

Since,

**Question 3: The value of** **is**

(a) π/9 (b) 5π/9

(c) -5π/9 (d) 7π/18

**Answer (d)**

**Explanation:**

**Question 4: Let** , Then value of θ is

(a) π/3 (b) π/2

(c) 2π/3 (d) – 2π/3

**Answer (a)**

**Explanation:**

**Question 5:** The principal value of is

(a) 2π/5 (b) -2π/5

(c) 3π/5 (d) -3π/5

**Answer: (b) -2π/5**

**Explanation:**

{since }

= –

= –2π/5

**Question 6:** is equal to:

(a) 1/2 (b) 1/3

(c) -1 (d) 1

**Answer: (d) 1**

**Explanation:**

= sin[π/3 – (-π/6)]

= sin(π/3 + π/6)

= sin (π/2)

= 1

**Question 7:** The domain of is

(a) [0, 1] (b) [– 1, 1]

(c) [-1/2, 1/2] (d) [–2, 2]

**Answer: (c) [-1/2, 1/2]**

**Explanation:** Let, .

2x = sin θ.

Since, – 1 ≤ sin θ ≤ 1

– 1 ≤ 2x ≤ 1\Rightarrow\sin^{-1}(2x)\sin^{–1}x + \sin^{–1}y = \pi/2\cos^{–1}x + \cos^{–1}y\sin^{–1} x + \sin^{–1} y = \pi/2\Rightarrow [(\pi/2) – \cos^{-1}x] + [(\pi/2) – \cos^{-1}y] = \pi/2\cos^{-1}x + \cos^{-1}y\cos^{-1}x + \cos^{-1}y\cos^{–1}x\cos^{-1}x\sin [\cot^{–1} (\cos (\tan^{–1} 1))]\sin [\cot^{–1} (\cos (\tan^{–1} 1))]\sin [\cot^{–1} (\cos (\tan^{–1} (\tan \pi/4))]= \sin[\cot^{-1} (\cos \pi/4)]= \sin[\cot^{-1}(1/√2)]= \sin [\sin^{-1}(√(⅔))]y = \cos^{–1} (x^2 – 4)y = \cos^{–1} (x^2 – 4 )\Rightarrow \cos y = x^2 – 4x^2 – 4x^22 \sin^{–1}x + \cos^{–1}x2 \sin^{–1}x + \cos^{–1}x\sin^{–1}\sin^{–1}x\sin^{–1}x + (\sin^{–1}x + \cos^{–1}x)2 \sin^{–1}x + \cos{–1}x2 \sin^{–1}x + \cos{–1}x\sin (2 \tan^{–1} (.75))\sin 1.5\sin (2\tan^{–1} (.75))\tan^{–1} (.75)\Rightarrow \tan \theta = 0.75\Rightarrow \tan \theta = 3/4\sin (2\tan^{–1} (.75)) = \sin 2θ\thetatan^{-1}(.75) = \theta\sin (2\tan^{–1} (.75)) = .96\sin(\tan^{-1} x)x/√(1 – x^2)1/√(1 – x^2)1/√(1 + x^2)x/√(1 + x^2)x/√(1 + x^2)\tan^{-1}x =\sin \theta = x/√(1 + x^2)\cos \theta = 1/√(1 + x^2)\sin(\tan^{-1} x) = \sin \theta = x/√(1 + x^2)2\sec^{-1}2+\sin^{-1}(\frac{1}{2})2\sec^{-1}2+\sin^{-1}(\frac{1}{2})= \frac{2\pi}{3} + \frac{\pi}{6}=\frac{5\pi}{6}\tan^2(\sec^{-1}2) +\cot^2(\opratorname{cosec}^{-1}3)f(x) =\sin^{-1}\sqrt{1-x}\cot[\cos^{-1}(\frac{7}{25})]\sin(\tan^{-1}x), |x|<1\frac{x}{\sqrt{1-x^2}}\frac{1}{\sqrt{1-x^2}}\frac{1}{\sqrt{1+x^2}}\frac{x}{\sqrt{1+x^2}}\cos^{-1}\alpha+\cos^{-1}\beta+\cos^{-1}\gamma = 3\pi\alpha(\beta +\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)\cos^{-1}x\sin^{-1}[\cos(33π/5)]\cos^{-1}(2x-1)\operatorname{cosec}^{-1}x\cos(\sin^{-1}\frac{3}{5}+\cos^{-1})= 0\cos^{-1}(2x^2-1),0 \leq x\lea 12\cos^{-1}x\sin^{-1}x\pi – 2\cos^{-1}x\pi + 2\cos^{-1}x\sqrt{1+\cos 2x} = \sqrt{2}\cos^{-1}(\cos x)2\cos^{-1}(\frac{-1}{2})+2\sin^{-1}(\frac{-1}{2})-\cos^{-1}(-1)\cot[\frac{1}{2}\sin^{-1}\frac{\sqrt{3}}{2}]\sin(\cot^{-1}x)\sqrt{1+x^2}(1+x^2)^{-3/2}(1+x^2)^{-1/2}$

**Answer (d)**