<s>[INST] How does $\cos\arcsin(\frac{3}{5})\cos\arctan(\frac{7}{24})-\sin\arcsin(\frac{3}{5})\sin\arctan\left(\frac{7}{24}\right)$ simplify to $\frac{3}{5}$? The question is to prove $\arcsin\left(\frac{3}{5}\right)+\arctan\left(\frac{7}{24}\right)=\arccos\left(\frac{3}{5}\right)$ which can be easily done by taking cos of both side and drawing triangles. However, the worked solutions does a simplification from LHS to RHS instead, namely $$\cos\arcsin\left(\frac{3}{5}\right)\cos\arctan\left(\frac{7}{24}\right)-\sin\arcsin\left(\frac{3}{5}\right)\sin\arctan\left(\frac{7}{24}\right)=\frac{3}{5}$$which I don't understand. Can someone please explain?
 [/INST] The provided solution first employs the angle addition identity $$\cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta,$$ with the choices $$\alpha = \arcsin \frac{3}{5}, \quad \beta = \arctan \frac{7}{24}.$$  That is to say, $$\sin \alpha = \frac{3}{5}, \quad \tan \beta = \frac{7}{24}.$$  In order to proceed with the evaluation, we now recall $$\sin^2 \theta + \cos^2 \theta = 1, \\ \tan^2 \theta + 1 = \sec^2 \theta.$$  So $$\frac{9}{25} = \sin^2 \alpha = 1 - \cos^2 \alpha,$$ hence $$\cos \alpha = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}.$$  Similarly, $$\frac{49}{24^2} = \tan^2 \beta = \sec^2 \beta - 1,$$ or $$\cos \beta = \frac{24}{25}.$$  It follows that $\sin \beta = \cos \beta \tan \beta = \frac{7}{25}$ and $$\cos(\alpha + \beta) = \frac{4}{5} \frac{24}{25} - \frac{3}{5} \frac{7}{25} = \frac{3}{5}.$$  This solution assumes that $-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2}$ and  $-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}$.
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