Question
Evaluate: $\sin\Big\{\cos^{-1}\Big(-\frac{3}{5}\Big)+\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$
✓
Answer
$\sin\Big\{\cos^{-1}\Big(-\frac{3}{5}\Big)+\cot^{-1}\Big(-\frac{5}{12}\Big)\Big\}$
$=\sin\Big\{\pi-\cos^{-1}\Big(\frac{3}{5}\Big)+\pi-\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$=\sin\Big\{2\pi-\Big[\cos^{-1}\Big(\frac{3}{5}\Big)+\cot^{-1}\Big(\frac{5}{12}\Big)\Big]\Big\}$
$=-\sin\Big\{\cos^{-1}\Big(\frac{3}{5}\Big)+\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$=-\sin\begin{Bmatrix}\sin^{-1}\Bigg[\sqrt{1-\Big(\frac{3}{5}\Big)^2}\Bigg]=\sin^{-1}\begin{bmatrix}\frac{\frac{12}{5}}{\sqrt{1+\big(\frac{12}{5}\big)^2}}\end{bmatrix}\end{Bmatrix}$
$=-\sin\Big(\sin^{-1}\frac{4}{5}+\sin^{-1}\frac{12}{13}\Big)$
$=-\sin\Bigg\{\sin^{-1}\Bigg[\frac{4}{5}\times\sqrt{1-\Big(\frac{12}{13}\Big)^2}+\frac{12}{13}\times\sqrt{1-\Big(\frac{4}{5}\Big)^2}\Bigg]\Bigg\}$
$=-\sin\Big[\sin^{-1}\Big(\frac{20}{65}+\frac{36}{64}\Big)\Big]$
$=-\sin\Big[\sin^{-1}\Big(\frac{56}{65}\Big)\Big]$
$=-\frac{56}{65}$
$=\sin\Big\{\pi-\cos^{-1}\Big(\frac{3}{5}\Big)+\pi-\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$=\sin\Big\{2\pi-\Big[\cos^{-1}\Big(\frac{3}{5}\Big)+\cot^{-1}\Big(\frac{5}{12}\Big)\Big]\Big\}$
$=-\sin\Big\{\cos^{-1}\Big(\frac{3}{5}\Big)+\cot^{-1}\Big(\frac{5}{12}\Big)\Big\}$
$=-\sin\begin{Bmatrix}\sin^{-1}\Bigg[\sqrt{1-\Big(\frac{3}{5}\Big)^2}\Bigg]=\sin^{-1}\begin{bmatrix}\frac{\frac{12}{5}}{\sqrt{1+\big(\frac{12}{5}\big)^2}}\end{bmatrix}\end{Bmatrix}$
$=-\sin\Big(\sin^{-1}\frac{4}{5}+\sin^{-1}\frac{12}{13}\Big)$
$=-\sin\Bigg\{\sin^{-1}\Bigg[\frac{4}{5}\times\sqrt{1-\Big(\frac{12}{13}\Big)^2}+\frac{12}{13}\times\sqrt{1-\Big(\frac{4}{5}\Big)^2}\Bigg]\Bigg\}$
$=-\sin\Big[\sin^{-1}\Big(\frac{20}{65}+\frac{36}{64}\Big)\Big]$
$=-\sin\Big[\sin^{-1}\Big(\frac{56}{65}\Big)\Big]$
$=-\frac{56}{65}$
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