JEE MAIN - Mathematics (2023 - 31st January Morning Shift - No. 3)
If $${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 < \alpha < 13$$, then $${\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )$$ is equal to :
16
$$\pi$$
16 $$-$$ 5$$\pi$$
0
Explanation
$\sin ^{-1}\left(\frac{\alpha}{17}\right)=-\cos ^{4}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{77}{36}\right)$
Let $\cos ^{-1}\left(\frac{4}{5}\right)=p$ and $\tan ^{-1}\left(\frac{77}{36}\right)=q$
$\Rightarrow \sin \left(\sin ^{-1} \frac{\alpha}{17}\right)=\sin (q-p)$
$=\sin q \cdot \cos p-\cos q \cdot \sin p$
$\Rightarrow \frac{\alpha}{17}=\frac{77}{85} \cdot \frac{4}{5}-\frac{36}{85} \cdot \frac{3}{5}$
$\Rightarrow \alpha=\frac{200}{25}=8$
$\sin ^{-1} \sin 8+\cos ^{-1} \cos 8$
$= -8+3 \pi+8-2 \pi$
$=\pi$
Let $\cos ^{-1}\left(\frac{4}{5}\right)=p$ and $\tan ^{-1}\left(\frac{77}{36}\right)=q$
$\Rightarrow \sin \left(\sin ^{-1} \frac{\alpha}{17}\right)=\sin (q-p)$
$=\sin q \cdot \cos p-\cos q \cdot \sin p$
$\Rightarrow \frac{\alpha}{17}=\frac{77}{85} \cdot \frac{4}{5}-\frac{36}{85} \cdot \frac{3}{5}$
$\Rightarrow \alpha=\frac{200}{25}=8$
$\sin ^{-1} \sin 8+\cos ^{-1} \cos 8$
$= -8+3 \pi+8-2 \pi$
$=\pi$
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