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?

As the inverse trigonometric functions are multi-valued with a respective principle value
or
as sine/cosine/tangent functions are periodic, it is incorrect to apply cosines in both sides.
For example,

*

*$x$ in general $\ne -x$  but $\cos(-x)=\cos(x)$

*$x$ in general $\ne \pi-x$  but $\sin(\pi-x)=\sin(x)$

*$x\ne\pi+x$  but $\tan(\pi+x)=\tan(x)$
Using the ranges of principle values of $\arctan,$ if $\arctan\dfrac7{24}=y$
$\implies0<y<\dfrac\pi2$
and $\tan y=\dfrac7{24}$
and $\sin y,\cos y>0$
Consequently, $\dfrac{\sin y}7=\dfrac{\cos y}{24}=+\sqrt{\dfrac{\sin^2y+\cos^2y}{7^2+24^2}}$
$\implies\arctan\dfrac7{24}=\arcsin\dfrac7{25}=\arccos\dfrac{24}{25}$
Now use Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $ in the left hand side
For the right hand side, if $\arccos\dfrac35=u$
$\implies0<u<\dfrac\pi2$ and $\cos u=\dfrac35$ and $\sin u>0$
$\implies\sin u=\sqrt{1-\cos^2u}=\dfrac45$
$\implies\arcsin\dfrac45=u=\arccos\dfrac35$
