Show that, $2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$ Show that, 
$$2\arctan\left(\frac{1}{3}\right)+2\arcsin\left(\frac{1}{5\sqrt2}\right)-\arctan\left(\frac{1}{7}\right)=\frac{\pi}{4}$$
There is a mixed of sin and tan, how  can I simplify this to $\frac{\pi}{4}$
We know the identity of $\arctan\left(\frac{1}{a}\right)+\arctan\left(\frac{1}{b}\right)=\arctan\left(\frac{a+b}{ab-1}\right)$
 A: From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),
$$\arctan x+\arctan y=\begin{cases} \arctan\dfrac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\dfrac{x+y}{1-xy} & \mbox{if } xy>1\end{cases} $$
$$\implies 2\arctan x=\arctan \frac{2x}{1-x^2}\text{  if }x^2<1$$
Finally use $$\arcsin x=\arctan\dfrac x{\sqrt{1-x^2}}$$
A: It would be easier to attack if you would substitute $\tan^{-1}{\frac17}\;$ for $\sin^{-1}{\frac1{5\sqrt2}}\;$.
A: Explanation for your question here
$2arctan(\frac{1}{3})$ = $arctan(\frac{3}{4})$ as you already understand
$arcsin(x) = arctan(\frac{x}{\sqrt{1-x^2}})$Inverse trigonometric functions
This simplifies $arcsin(\frac{1}{5\sqrt2}) = arctan (\frac{1}{7})$ which makes 
$2arcsin(\frac{1}{5\sqrt2}) = 2arctan(\frac{1}{7})$
Now the solution is
$2arctan(\frac{1}{3})+ 2arcsin(\frac{1}{5\sqrt2})-arctan(\frac{1}{7})$
simplifies into
$arctan(\frac{3}{4})+ 2arctan(\frac{1}{7})-arctan(\frac{1}{7})$
$arctan(\frac{3}{4})+arctan(\frac{1}{7})$ 
$arctan(\frac{(21+4)}{(28-3)}) = arctan(1) = \frac{\pi}{4}$
