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JEE Main 2019
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Medium

Question

cos(sin135+sin1513+sin13365)\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right) is equal to:

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Solution

cos(sin135+sin1513+sin13365)cos(tan134+tan1512+tan13356)cos(tan1(34+5121+34512)+tan13356)cos(tan15633+cot15633)cos(π2)=0\begin{aligned} & \cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right) \\ & \cos \left(\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{33}{56}\right) \\ & \cos \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{5}{12}}{1+\frac{3}{4} \cdot \frac{5}{12}}\right)+\tan ^{-1} \frac{33}{56}\right) \\ & \cos \left(\tan ^{-1} \frac{56}{33}+\cot ^{-1} \frac{56}{33}\right) \\ & \cos \left(\frac{\pi}{2}\right)=0 \end{aligned}

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