The value of sin^-1 (cos x), \(0\le x\le\pi\) is

(a)

\(\pi-x\)

(b)

\(x-\frac{\pi}{2}\)

(c)

\(\frac{\pi}{2}-x\)

(d)

\(x-\pi\)

If \(\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}\); then cos^-1 x + cos^-1 y is equal to

(a)

\(\frac{2\pi}{3}\)

(b)

\(\frac{\pi}{3}\)

(c)

\(\frac{\pi}{6}\)

(d)

\(\pi\)

\(\sin ^{-1} \frac{3}{5}-\cos ^{-1} \frac{12}{13}+\sec ^{-1} \frac{5}{3}-\operatorname{cosec}^{-1} \frac{13}{12}\) is equal to

(a)

2\(\pi\)

(b)

\(\pi\)

(c)

0

(d)

tan^-1\(\frac{12}{65}\)

If sin⁻¹x = 2sin⁻¹ \(\alpha\) has a solution, then

(a)

\(|\alpha |\le \frac { 1 }{ \sqrt { 2 } } \)

(b)

\(|\alpha |\ge \frac { 1 }{ \sqrt { 2 } } \)

(c)

\(|\alpha |<\frac { 1 }{ \sqrt { 2 } } \)

(d)

\(|\alpha |>\frac { 1 }{ \sqrt { 2 } } \)

\(\sin ^{-1}(\cos x)=\frac{\pi}{2}-x\) is valid for

(a)

\(-\pi \le x\le 0\)

(b)

\(0 \le x\le \pi\)

(c)

\(-\frac { \pi }{ 2 } \le x\le \frac { \pi }{ 2 } \)

(d)

\(-\frac { \pi }{ 4 } \le x\le \frac { 3\pi }{ 4 } \)

If sin^-1 x+sin^-1 y+sin^-1 \(z = \frac{3\pi}{2}\), the value of x²⁰¹⁷+y²⁰¹⁸+z²⁰¹⁹\(-\frac { 9 }{ { x }^{ 101 }+{ y }^{ 101 }+{ z }^{ 101 } } \)is

(a)

0

(b)

1

(c)

2

(d)

3

If \(\cot ^{-1} x=\frac{2 \pi}{5}\) for some x \(\in\) R, the value of tan^-1 x is

(a)

\(-\frac{\pi}{10}\)

(b)

\(\frac{\pi}{5}\)

(c)

\(\frac{\pi}{10}\)

(d)

\(-\frac{\pi}{5}\)

The domain of the function defined by \(f(x)=\sin ^{-1} \sqrt{x-1}\) is

(a)

[1, 2]

(b)

[-1, 1]

(c)

[0, 1]

(d)

[-1, 0]

If \(x = \frac{1}{5}\), the value of cos (cos^-1x+2sin^-1x) is

(a)

\(-\sqrt { \frac { 24 }{ 25 } } \)

(b)

\(\sqrt { \frac { 24 }{ 25 } } \)

(c)

\(\frac{1}{5}\)

(d)

\(-\frac{1}{5}\)

\(\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)\) is equal to

(a)

\(\frac { 1 }{ 2 } \ { cos }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)

(b)

\(\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)

(c)

\(\frac { 1 }{ 2 } {tan }^{ -1 }\left( \frac { 3 }{ 5 } \right) \)

(d)

\({ tan}^{ -1 }\left( \frac { 1}{ 2 } \right) \)

If the function f(x) = sin^-1(x² - 3), then x belongs to

(a)

[-1, 1]

(b)

[\(\sqrt2\), 2]

(c)

\(\\ \\ \\ \left[ -2,-\sqrt { 2 } \right] \cup \left[ \sqrt { 2 } ,2 \right] \)

(d)

\([-2,-\sqrt{2}]\)

If cot^-1 2 and cot^-1 3 are two angles of a triangle, then the third angle is

(a)

\(\frac{\pi}{4}\)

(b)

\(\frac{3\pi}{4}\)

(c)

\(\frac{\pi}{6}\)

(d)

\(\frac{\pi}{3}\)

\(\sin ^{-1}\left(\tan \frac{\pi}{4}\right)-\sin ^{-1}\left(\sqrt{\frac{3}{x}}\right)=\frac{\pi}{6}\). Then x is a root of the equation

(a)

x²−x−6 = 0

(b)

x²−x−12 = 0

(c)

x²+x−12 = 0

(d)

x²+x−6 = 0

sin^-1(2cos²x-1)+cos^-1(1-2sin²x)=

(a)

\(\frac{\pi}{2}\)

(b)

\(\frac{\pi}{3}\)

(c)

\(\frac{\pi}{4}\)

(d)

\(\frac{\pi}{6}\)

If \(\cot ^{-1}(\sqrt{\sin \alpha})+\tan ^{-1}(\sqrt{\sin \alpha})=u\), then cos2u is equal to

(a)

tan²\(\alpha\)

(b)

0

(c)

-1

(d)

tan2\(\alpha\)

If |x| \(\le\) 1, then 2 tan^-1 x-sin^-1\(\frac{2x}{1+x^2}\) is equal to

(a)

tan^-1x

(b)

sin^-1x

(c)

0

(d)

\(\pi\)

The equation \(\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)\)has

(a)

no solution

(b)

unique solution

(c)

two solutions

(d)

infinite number of solutions

If \(\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}\), then x is equal to

(a)

\(\frac{1}{2}\)

(b)

\(\frac{1}{\sqrt{5}}\)

(c)

\(\frac{2}{\sqrt{5}}\)

(d)

\(\frac{\sqrt3}{2}\)

If \(\sin ^{-1} \frac{x}{5}+\operatorname{cosec}^{-1} \frac{5}{4}=\frac{\pi}{2}\), then the value of x is

(a)

4

(b)

5

(c)

2

(d)

3

sin (tan^-1x), |x| < 1 is equal to

(a)

\(\frac{x}{\sqrt{1-x^2}}\)

(b)

\(\frac{1}{\sqrt{1-x^2}}\)

(c)

\(\frac{1}{\sqrt{1+x^2}}\)

(d)

\(\frac{x}{\sqrt{1+x^2}}\)