The general solution satisfying both the equations $\sin x=-\frac{3}{5}$ and $\cos x=-\frac{4}{5}$ is
$x=(2 n+1) \pi+\tan ^{-1}\left(\frac{3}{4}\right), n \in Z$
$x=2 n \pi+\tan ^{-1}\left(\frac{3}{4}\right), n \in Z$
$x=n \pi+\tan ^{-1}\left(\frac{3}{4}\right), n \in Z$
$x=n \pi \pm \tan ^{-1}\left(\frac{3}{4}\right), n \in Z$
The number of solution of $\tan ^{-1} 1+\frac{1}{2} \cos ^{-1} x^2-\tan ^{-1} \left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=0$ is
3
0
1
infinitely many
$$ \tanh ^{-1}(\sin \theta)= $$
$\sinh ^{-1}(\operatorname{cosec} \theta)$
$\sinh ^{-1}(\sec \theta)$
$\cosh ^{-1}(\operatorname{cosec} \theta)$
$\cosh ^{-1}(\sec \theta)$
In $\triangle A B C$, if $a=8, b=10, c=12$, then $\frac{r}{R}=$
$\frac{8}{15}$
$\frac{7}{16}$
$\frac{3}{5}$
$\frac{5}{8}$
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