# Multiple choice (Inverse Trigonometry)

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

Question 1: The value ofÂ  corresponding to principal branches is

(a) -Ï€/12Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (b) 0

(c) Ï€Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (d) Ï€/3

Explanation:

Question 2: The value of is

(a) Â  Â  Â  Â  Â  Â (b)Â

(c) 1/xÂ  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (d)

Explanation:

Let

Since,

Question 3: The value of is

(a) Ï€/9Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (b) 5Ï€/9

(c) -5Ï€/9Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (d)Â  7Ï€/18

Explanation:

Question 4: Let , Then value of Î¸ is

(a) Ï€/3Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (b) Ï€/2

(c) 2Ï€/3Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (d) – 2Ï€/3

Explanation:

Question 5: The principal value of is

(a) 2Ï€/5Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (b) -2Ï€/5

(c) 3Ï€/5Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  (d) -3Ï€/5

Explanation:

{since }

= â€“

= â€“2Ï€/5

Question 6: is equal to:

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

(c) -1Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (d) 1

Explanation:

= sin[Ï€/3 â€“ (-Ï€/6)]

= sin(Ï€/3 + Ï€/6)

= sin (Ï€/2)

= 1

Question 7: The domain of is

(a) [0, 1]Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â  Â (b) [â€“ 1, 1]

(c) [-1/2, 1/2]Â  Â  Â  Â  Â  Â  Â  Â  (d) [â€“2, 2]

Explanation: Let, .

2x = sin Î¸.

Since, â€“ 1 â‰¤ sin Î¸ â‰¤ 1

â€“ 1 â‰¤ 2x â‰¤ 1\Rightarrow\sin^{-1}(2x)\sin^{â€“1}x + \sin^{â€“1}y = \pi/2\cos^{â€“1}x + \cos^{â€“1}y\sin^{â€“1} x + \sin^{â€“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^{â€“1}x\cos^{-1}x\sin [\cot^{â€“1} (\cos (\tan^{â€“1} 1))]\sin [\cot^{â€“1} (\cos (\tan^{â€“1} 1))]\sin [\cot^{â€“1} (\cos (\tan^{â€“1} (\tan \pi/4))]= \sin[\cot^{-1} (\cos \pi/4)]= \sin[\cot^{-1}(1/âˆš2)]= \sin [\sin^{-1}(âˆš(â…”))]y = \cos^{â€“1} (x^2 â€“ 4)y = \cos^{â€“1} (x^2 â€“ 4 )\Rightarrow \cos y = x^2 â€“ 4x^2 â€“ 4x^22 \sin^{â€“1}x + \cos^{â€“1}x2 \sin^{â€“1}x + \cos^{â€“1}x\sin^{â€“1}\sin^{â€“1}x\sin^{â€“1}x + (\sin^{â€“1}x + \cos^{â€“1}x)2 \sin^{â€“1}x + \cos{â€“1}x2 \sin^{â€“1}x + \cos{â€“1}x\sin (2 \tan^{â€“1} (.75))\sin 1.5\sin (2\tan^{â€“1} (.75))\tan^{â€“1} (.75)\Rightarrow \tan \theta = 0.75\Rightarrow \tan \theta = 3/4\sin (2\tan^{â€“1} (.75)) = \sin 2Î¸\thetatan^{-1}(.75) = \theta\sin (2\tan^{â€“1} (.75)) = .96\sin(\tan^{-1} x)x/âˆš(1 â€“ x^2)1/âˆš(1 â€“ x^2)1/âˆš(1 + x^2)x/âˆš(1 + x^2)x/âˆš(1 + x^2)\tan^{-1}x =\sin \theta = x/âˆš(1 + x^2)\cos \theta = 1/âˆš(1 + x^2)\sin(\tan^{-1} x) = \sin \theta = x/âˆš(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(33Ï€/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}\$