Hello, prasum!
$\displaystyle \text}Find the value of: }\:169e^{\left(\pi+\sin^{\text{-}1}\!\frac{12}{13}+\cos^{\text{-}1}\!\frac{5}{13}\right)}$
$\displaystyle \text{Examine that exponent: }\:\pi + \underbrace{\sin^{\text{-}1}\!\!\left(\tfrac{12}{13}\right)}_{\alpha} + \underbrace{\cos^{\text{-}1}\!\!\left(\tfrac{5}{13}\right)}_{\beta} $
$\displaystyle \text{We have: }\:\begin{Bmatrix}\alpha \:=\:\sin^{-1}\frac{12}{13} & \Rightarrow & \sin\alpha \:=\:\frac{12}{13} \\ \\[-3mm]
\beta \:=\:\cos^{-1}\frac{5}{13} & \Rightarrow & \cos\beta \:=\:\frac{5}{12} \end{Bmatrix}$
Note that $\displaystyle \alpha$ and $\displaystyle \beta$ are the same angle!
Code:
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5
Hence, the exponent is: .$\displaystyle \pi + \alpha + \alpha \:=\:\pi + 2\sin^{\text{-}1}\frac{12}{13}$
But what is the point of all this?
We could have cranked out the answer on our calculator.