\(\cos\left(\sin^{-1}\left(\dfrac{3}{5}\right)+\sin^{-1}\left(\dfrac{5}{13}\right)+\sin^{-1}\left(\dfrac{33}{65}\right)\right)\) is equal to:
0
Let \(A=\sin^{-1}\dfrac{3}{5}\), so \(\sin A=\dfrac{3}{5}\), \(\cos A=\dfrac{4}{5}\).
Let \(B=\sin^{-1}\dfrac{5}{13}\), so \(\sin B=\dfrac{5}{13}\), \(\cos B=\dfrac{12}{13}\).
Let \(C=\sin^{-1}\dfrac{33}{65}\), so \(\sin C=\dfrac{33}{65}\), and \(\cos C=\sqrt{1-\left(\dfrac{33}{65}\right)^2}=\sqrt{\dfrac{4225-1089}{4225}}=\dfrac{56}{65}\).
Compute \(\cos(A+B)=\cos A\cos B-\sin A\sin B=\dfrac{4}{5}\cdot\dfrac{12}{13}-\dfrac{3}{5}\cdot\dfrac{5}{13}=\dfrac{48-15}{65}=\dfrac{33}{65}\).
Compute \(\sin(A+B)=\sin A\cos B+\cos A\sin B=\dfrac{3}{5}\cdot\dfrac{12}{13}+\dfrac{4}{5}\cdot\dfrac{5}{13}=\dfrac{36+20}{65}=\dfrac{56}{65}\).
Observe that \(\cos(A+B)=\dfrac{33}{65}=\sin C\) and \(\sin(A+B)=\dfrac{56}{65}=\cos C\), which means \(A+B=\dfrac{\pi}{2}-C\), i.e. \(A+B+C=\dfrac{\pi}{2}\).
Therefore \(\cos(A+B+C)=\cos\dfrac{\pi}{2}=0\).
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1. There exists \({\rm{\theta }} \in \left( { - \frac{{\rm{\pi }}}{2},\frac{{\rm{\pi }}}{2}} \right)\) for which tan -1 (tan θ) ≠ θ
2. \({\sin ^{ - 1}}\left( {\frac{1}{3}} \right) - {\sin ^{ - 1}}\left( {\frac{1}{5}} \right) = {\sin ^{ - 1}}\left( {\frac{{2\sqrt 2 \left( {\sqrt 3 - 1} \right)}}{{15}}} \right)\)
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1. \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\rm{\pi }}\)
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Which of the above statements is/are correct?The value of \({\rm{tan}}\left( {2{{\tan }^{ - 1}}\frac{1}{5} - \frac{\pi }{4}} \right)\) is
