Trigonometry

If $\alpha,-\frac{\pi}{2}<\alpha<\frac{\pi}{2}$ is the solution of $4 \cos \theta+5 \sin \theta=1$, then the value of $\tan \alpha$ is

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy

${\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}x$ is equal to :

Let M and m respectively be the maximum and minimum values of the function
f(x) = tan^$-$1 (sin x + cos x) in $\left[ {0,{\pi \over 2}} \right]$, then the value of tan(M $-$ m) is equal to :

Consider two geometric progressions $a_1,a_2,a_3\mathrm{.......}{a}_n$ & $b_1, b_2, b_3,\mathrm{.....}{b}_n$ with $a_r=\frac{1}{b_r}=2^{r-1}$ and another sequence $t_1,t_2,t_3\mathrm{.......}{t}_n$ such that $t_r={\cot }^{-1}\left(2a_r+b_r\right)$ then $\underset{n\to \infty }{\lim }\sum _{r=1}^n{t}_r$ is -

A possible value of $\tan \left( {{1 \over 4}{{\sin }^{ - 1}}{{\sqrt {63} } \over 8}} \right)$ is :

The value $\tan \left.{\sin }^{-1}\left(\cos \left({\sin }^{-1}x\right)\right)\right\}\tan \left.{\cos }^{-1}\left(\sin \left({\cos }^{-1}x\right)\right)\right\}$  is equal to (where $0<x<1$ ) -

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$ is :

2$\pi$ - $\left( {{{\sin }^{ - 1}}{4 \over 5} + {{\sin }^{ - 1}}{5 \over {13}} + {{\sin }^{ - 1}}{{16} \over {65}}} \right)$ is equal to :

If $\frac{\sin^1 x}{a} = \frac{\cos^{-1} x}{b} = \frac{\tan^{-1} y}{c}$; $0 < x < 1$, then the value of $\cos\left(\frac{\pi c}{a + b}\right)$ is:

The value of ${\sin ^{ - 1}}\left( {{{12} \over {13}}} \right) - {\sin ^{ - 1}}\left( {{3 \over 5}} \right)$ is equal to :