Question

Prove that: $\tan \left(\frac{\pi }{4}+x\right)+\tan \left(\frac{\pi }{4}-x\right)=2\sec 2x$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\tan \left(\frac{\mathrm{\pi }}{4}+x\right)+\tan \left(\frac{\mathrm{\pi }}{4}-x\right) \\ =\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan x}{1-\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan x}+\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan x}{1+\tan {\displaystyle \frac{\mathrm{\pi }}{4}}\tan x}\left[\because \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}\mathrm{and}\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A\tan B}\right] \end{gathered}$$

$$\begin{gathered} =\frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x} \\ =\frac{{\left(1+\tan x\right)}^2+{\left(1-\tan x\right)}^2}{\left(1+\tan x\right)\left(1-\tan x\right)} \\ =\frac{2(1+{\tan }^{2}x)}{\left(1-{\tan }^{2}x\right)}=\frac{2\left({\sec }^{2}x\right)}{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}} \end{gathered}$$

$$\begin{gathered} =\frac{2\left({\sec }^{2}x\right)\left({\cos }^{2}x\right)}{\cos 2x}\left(\because {\cos }^{2}x-{\sin }^{2}x=\cos 2x\right) \\ =\frac{2\times 1}{\cos 2x} \\ =2\sec 2x=\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$