JEE Main
Exam JEE Main
Subject Mathematics
Chapter Inverse Trigonometric Functions JEE Main · Mathematics · Inverse Trigonometric Functions
What is the value of ^ -1 (12)/(13)+ ^ -1 (4)/(5)+ ^ -1 (63)/(16)...
What is the value of
sin − 1 12 13 + cos − 1 4 5 + tan − 1 63 16 \sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{63}{16} sin − 1 13 12 + cos − 1 5 4 + tan − 1 16 63 is equal to
Options A π \pi π B π 2 \frac{\pi}{2} 2 π C π 6 \frac{\pi}{6} 6 π D 3 π 4 \frac{3\pi}{4} 4 3 π
Step-by-step solution
sin − 1 ( 12 13 ) + cos − 1 ( 4 5 ) + tan − 1 ( 63 16 ) \sin ^{-1}\left(\frac{12}{13}\right)+\cos ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right) sin − 1 ( 13 12 ) + cos − 1 ( 5 4 ) + tan − 1 ( 16 63 ) = sin − 1 ( 12 13 ) + π 2 − sin − 1 ( 4 5 ) + tan − 1 ( 63 16 ) { ∵ sin − 1 x + cos − 1 x = π 2 } = sin − 1 ( 12 13 ) − sin − 1 ( 4 5 ) + π 2 + sin − 1 ( 63 ( 63 ) 2 + ( 16 ) 2 ) { ∵ sin − 1 x ± sin − 1 y = sin − 1 ( x 1 − y 2 ± y 1 − x 2 ) } = sin − 1 { 12 13 ⋅ 1 − ( 4 5 ) 2 − 4 5 ⋅ 1 − ( 12 13 ) 2 } + π 2 + sin − 1 ( 63 65 ) \begin{array}{r}=\sin ^{-1}\left(\frac{12}{13}\right)+\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{5}\right)+\tan ^{-1}\left(\frac{63}{16}\right) \\ \left\{\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right\} \\ =\sin ^{-1}\left(\frac{12}{13}\right)-\sin ^{-1}\left(\frac{4}{5}\right)+\frac{\pi}{2}+\sin ^{-1}\left(\frac{63}{\sqrt{(63)^2+(16)^2}}\right) \\ \quad\left\{\because \sin ^{-1} x \pm \sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^2} \pm y \sqrt{\left.1-x^2\right)}\right\}\right. \\ =\sin ^{-1}\left\{\frac{12}{13} \cdot \sqrt{1-\left(\frac{4}{5}\right)^2}-\frac{4}{5} \cdot \sqrt{1-\left(\frac{12}{13}\right)^2}\right\} \\ +\frac{\pi}{2}+\sin ^{-1}\left(\frac{63}{65}\right)\end{array} = sin − 1 ( 13 12 ) + 2 π − sin − 1 ( 5 4 ) + tan − 1 ( 16 63 ) { ∵ sin − 1 x + cos − 1 x = 2 π } = sin − 1 ( 13 12 ) − sin − 1 ( 5 4 ) + 2 π + sin − 1 ( ( 63 ) 2 + ( 16 ) 2 63 ) { ∵ sin − 1 x ± sin − 1 y = sin − 1 ( x 1 − y 2 ± y 1 − x 2 ) } = sin − 1 { 13 12 ⋅ 1 − ( 5 4 ) 2 − 5 4 ⋅ 1 − ( 13 12 ) 2 } + 2 π + sin − 1 ( 65 63 ) = sin − 1 { 12 13 ⋅ 3 5 − 4 5 ⋅ 5 13 } + π 2 + sin − 1 ( 63 65 ) and = sin − 1 ( 16 65 ) + sin − 1 ( 63 65 ) + π 2 and = sin − 1 { 16 65 ⋅ 1 − ( 63 65 ) 2 + 63 65 ⋅ 1 − ( 16 65 ) 2 } + π 2 and = sin − 1 ( 1 ) + π 2 ⇒ π 2 + π 2 = π \begin{aligned}& =\sin ^{-1}\left\{\frac{12}{13} \cdot \frac{3}{5}-\frac{4}{5} \cdot \frac{5}{13}\right\}+\frac{\pi}{2}+\sin ^{-1} \left(\frac{63}{65}\right)\\ \text{ and } =\sin ^{-1}\left(\frac{16}{65}\right)+\sin ^{-1}\left(\frac{63}{65}\right)+\frac{\pi}{2}\\ \text{ and } =\sin ^{-1}\left\{\frac{16}{65} \cdot \sqrt{1-\left(\frac{63}{65}\right)^2}+\frac{63}{65} \cdot \sqrt{1-\left(\frac{16}{65}\right)^2}\right\}+\frac{\pi}{2}\\ \text{ and } =\sin ^{-1}(1)+\frac{\pi}{2} \Rightarrow \frac{\pi}{2}+\frac{\pi}{2}=\pi\end{aligned} and = sin − 1 ( 65 16 ) + sin − 1 ( 65 63 ) + 2 π and = sin − 1 ⎩ ⎨ ⎧ 65 16 ⋅ 1 − ( 65 63 ) 2 + 65 63 ⋅ 1 − ( 65 16 ) 2 ⎭ ⎬ ⎫ + 2 π and = sin − 1 ( 1 ) + 2 π ⇒ 2 π + 2 π = π = sin − 1 { 13 12 ⋅ 5 3 − 5 4 ⋅ 13 5 } + 2 π + sin − 1 ( 65 63 ) $$$
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