Question

If $\tan \left(\pi /4+x\right)+\tan \left(\pi /4-x\right)=\lambda \sec 2x,\mathrm{then}$
(a) 3
(b) 4
(c) 1
(d) 2

Answer 1

(d) 2

$$\begin{gathered} \mathrm{Given}: \\ \tan \left(\frac{\mathrm{\pi }}{4}+x\right)+\tan \left(\frac{\mathrm{\pi }}{4}-x\right)=\mathrm{\lambda }\sec 2x \\ \Rightarrow \frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}+\tan x}{1-\tan \frac{\mathrm{\pi }}{4}\times \tan x}+\frac{\tan {\displaystyle \frac{\mathrm{\pi }}{4}}-\tan x}{1+\tan \frac{\mathrm{\pi }}{4}\times \tan x}=\mathrm{\lambda }\sec 2x \\ \Rightarrow \frac{1+\tan x}{1-\tan x}+\frac{1-\tan x}{1+\tan x}=\mathrm{\lambda }\sec 2x \\ \Rightarrow \frac{{\left(1+\tan x\right)}^2+{\left(1-\tan x\right)}^2}{\left(1-\tan x\right)\left(1+\tan x\right)}=\mathrm{\lambda }\sec 2x \\ \Rightarrow \frac{2\left(1+{\tan }^{2}x\right)}{1-{\tan }^{2}x}=\mathrm{\lambda }\sec 2x \end{gathered}$$

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