$\pi + \left(\sin^{-1} \frac{4}{5} + \sin^{-1} \frac{5}{13} + \sin^{-1} \frac{16}{65}\right)$ is equal to

  • A
    $\frac{\pi}{2}$
  • B
    $\frac{5\pi}{4}$
  • C
    $\frac{3\pi}{2}$
  • D
    $\frac{7\pi}{4}$

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Similar Questions

Let $E_1 = \{x \in R : x \neq 1 \text{ and } \frac{x}{x-1} > 0\}$ and $E_2 = \{x \in E_1 : \sin^{-1}(\log_e(\frac{x}{x-1})) \text{ is a real number}\}$. (Here,the inverse trigonometric function $\sin^{-1} x$ assumes values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$). Let $f : E_1 \rightarrow R$ be the function defined by $f(x) = \log_e(\frac{x}{x-1})$ and $g : E_2 \rightarrow R$ be the function defined by $g(x) = \sin^{-1}(\log_e(\frac{x}{x-1}))$. Match the items in $LIST I$ with $LIST II$.
$LIST I$ $LIST II$
$P$. The range of $f$ is $1$. $(-\infty, \frac{1}{1-e}] \cup [\frac{e}{e-1}, \infty)$
$Q$. The range of $g$ contains $2$. $(0, 1)$
$R$. The domain of $f$ contains $3$. $[-\frac{1}{2}, \frac{1}{2}]$
$S$. The domain of $g$ is $4$. $(-\infty, 0) \cup (0, \infty)$
$5$. $(-\infty, \frac{e}{e-1}]$
$6$. $(-\infty, 0) \cup (\frac{1}{2}, \frac{e}{e-1}]$

Let $x = \sin(2 \tan^{-1} \alpha)$ and $y = \sin(\frac{1}{2} \tan^{-1} \frac{4}{3})$. If $S = \{\alpha \in R : y^2 = 1 - x\}$,then $\sum_{\alpha \in S} 16 \alpha^3$ is equal to $...........$

The number of solutions of the equation $2\tan^{-1}(\cos^2 x) = \tan^{-1}(2\csc^2 x)$ in the interval $[0, 5\pi]$ is $m$. Then which of the following is true?

Let $(x, y)$ be such that $\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$. Match the statements in Column $I$ with the statements in Column $II$.
Column $I$ Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$ $(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$ $(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$ $(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$ $(s)$ lies on $(4x^2-1)(y^2-1)=0$

If $a^2+b^2+c^2=r^2$,then the value of $\tan ^{-1}\left(\frac{a b}{c r}\right)+\tan ^{-1}\left(\frac{b c}{a r}\right)+\tan ^{-1}\left(\frac{c a}{b r}\right)=$

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