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Inverse Trigonometric Functions

Multiple Choice Questions

241.

$\cos^{- 1} (\frac{1}{2}) + 2 \sin^{- 1} (\frac{1}{2})$ is equal to :

$\frac{π}{4}$
$\frac{π}{6}$
$\frac{π}{3}$
$\frac{2 π}{3}$

242.

A particle possess two velocities simultaneously at an angle of $\tan^{- 1} (\frac{12}{5})$ ; to each other. Their resultant is 15 m/s. If one velocity is 13 m/s, then the other will be :

5 m/s
4 m/s
12 m/s
13 m/s

243.

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

244.
If $\sin^{- 1} (\frac{x}{5}) + \csc^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then x is equal to

1

4

3

5

$We have, \sin^{- 1} (\frac{x}{5}) + \csc^{- 1} (\frac{5}{4}) = \frac{π}{2} \Rightarrow \sin^{- 1} (\frac{x}{5}) + \csc^{- 1} (\frac{5}{4}) = \frac{π}{2} \Rightarrow \sin^{- 1} [\frac{x}{5} \times \sqrt{1 - {(\frac{4}{5})}^{2}} + \frac{4}{5} \sqrt{1 - {(\frac{x}{5})}^{2}}] = \frac{π}{2} or \frac{x}{5} \times \frac{3}{5} + \frac{4}{5} \times \frac{\sqrt{25 - x^{2}}}{5} = \sin (\frac{π}{2}) \Rightarrow 3 x + 4 \sqrt{25 - x^{2}} = 25 \Rightarrow 4 \sqrt{25 - x^{2}} = 25 - 3 x$

On squaring both sides, we get

16(25 - x²) = 625 + 9x² - 150x

$\Rightarrow 4500 - 16 x^{2} = 625 + 9 x^{2} - 150 x \Rightarrow 16 x^{2} + 9 x^{2} - 150 x + 625 - 400 = 0 \Rightarrow 25 x^{2} - 150 x + 225 = 0 \Rightarrow x^{2} - 6 x + 9 = 0 or x^{2} - 3 x - 3 x + 9 = 0 x (x - 3) - 3 (x - 3) = 0 {(x - 3)}^{2} = 0 ∴ x = 3$

245.

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

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

246.

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

247.

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

248.

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

249.

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

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

250.

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 :

$\frac{π}{6}$
$\frac{π}{4}$
$\frac{π}{3}$
$\frac{π}{2}$

Previous Year Papers

Test Series

Test Yourself

Multiple Choice Questions

244. If sin-1x5 + csc-154 = π2, then x is equal to 1 4 3 5

244.
If $\sin^{- 1} (\frac{x}{5}) + \csc^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then x is equal to

1

4

3

5