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Inverse Trigonometric Functions

Multiple Choice Questions

71.

If $\sin^{- 1} (\frac{5}{x}) + \sin^{- 1} (\frac{12}{x}) = \frac{π}{2}$ , then x is equal to :

$\frac{7}{13}$
$\frac{4}{3}$
13
$\frac{13}{7}$

72.
The sum of the series $\tan^{- 1} (\frac{1}{1 + 1 + 1^{2}}) + \tan^{- 1} (\frac{1}{2 + 2 + 2^{2}}) + \tan^{- 1} (\frac{1}{3 + 3 + 3^{2}}) + . . . \infty$ is equal to :

$\frac{π}{4}$

$\frac{π}{2}$

$\frac{π}{3}$

$\frac{π}{6}$

$\frac{π}{2}$

Given series can be rewritten as

$\sum_{r = 1}^{\infty} \tan^{- 1} (\frac{1}{1 + r + r^{2}}) Now, \tan^{- 1} (\frac{1}{1 + r + r^{2}}) = \tan^{- 1} (\frac{r + 1 - r}{1 + r (r + 1)}) = \tan^{- 1} (r + 1) = \tan^{- 1} (r) ∴ \sum_{r = 0}^{n} [\tan^{- 1} (r + 1) - \tan^{- 1} (r)] = \tan^{- 1} (n + 1) - \tan^{- 1} (0) = \tan^{- 1} (n + 1) \Rightarrow \sum_{r = 0}^{\infty} \tan^{- 1} (\frac{1}{1 + r + r^{2}}) = \tan^{- 1} (\infty) = \frac{π}{2}$

73.

The avlue of $\sec [\tan^{- 1} (\frac{b + a}{b - a}) - \tan^{- 1} (\frac{a}{b})]$ is :

2
$\sqrt{2}$
4
1

74.

The regression coefficient of y on x is 2/3 and that of x on y is 4/3. The acute angle between the two regression lines is tan^-1(k), where k is equal to :