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Inverse Trigonometric Functions

Multiple Choice Questions

111.

If $\cos^{- 1} (p) + \cos^{- 1} (q) + \cos^{- 1} (r) = π$ , then p² + q² + r² + 2pqr is equal to

112.

If $\sin^{- 1} (\frac{x}{5}) + \csc^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then x is equal to

113.

$\sin (\frac{1}{2} \cos^{- 1} (\frac{4}{5}))$

$- \frac{1}{\sqrt{10}}$
$\frac{1}{\sqrt{10}}$
$- \frac{1}{10}$
$\frac{1}{10}$

114.

If $\cos^{- 1} (x) + \cos^{- 1} (y) + \cos^{- 1} (z) = 3 π$ , then xy + yz + zx is equal to

115.
If $\tan^{- 1} (\frac{x + 1}{x - 1}) + \tan^{- 1} (\frac{x - 1}{x}) = \tan^{- 1} (- 7)$ , then the value of x is

zero

- 2

1

2

$\tan^{- 1} (\frac{x + 1}{x - 1}) + \tan^{- 1} (\frac{x - 1}{x}) = \tan^{- 1} (- 7) \Rightarrow \frac{\frac{x + 1}{x - 1} + \frac{x - 1}{x}}{1 - \frac{(x + 1) (x - 1)}{x (x - 1)}} = (- 7) \Rightarrow \frac{x (x + 1) + {(x - 1)}^{2}}{x (x - 1) - (x^{2} - 1)} = - 7 \Rightarrow \frac{x^{2} + x + x^{2} + 1 - 2 x}{x^{2} - x - x^{2} + 1} = - 7 \Rightarrow \frac{2 x^{2} - x + 1}{1 - x} = \frac{- 7}{1} \Rightarrow - 7 + 7 x = 2 x^{2} - x + 1 \Rightarrow 2 x^{2} - 8 x + 8 = 0 \Rightarrow x^{2} - 4 x + 4 = 0 \Rightarrow {(x - 2)}^{2} = 0 \Rightarrow x = 2$

116.

If $\cos^{- 1} (\sqrt{p}) + \cos^{- 1} (\sqrt{1 - p}) + \cos^{- 1} (\sqrt{1 - q}) = \frac{3 π}{4}$ , then the value of q is

$\frac{2}{\sqrt{2}}$
1
1/2
1/3

117.

The soluton of $\sin^{- 1} (x) - \sin^{- 1} (2 x) = \pm \frac{π}{3}$ is

$\pm \frac{1}{3}$
$\pm \frac{1}{4}$
$\pm \frac{\sqrt{3}}{2}$
$\pm \frac{1}{2}$

118.

If $\cos^{- 1} (x) = α, (0 < x < 1)$ and $\sin^{- 1} (2 x \sqrt{1 - x^{2}})$ + $\sec^{- 1} (\frac{1}{2 x^{2} - 1}) = \frac{2 π}{3}$ , then $\tan^{- 1} (2 x)$ equals :