Question

If tan (π/4 + x) + tan (π/4 − x) = a, then tan² (π/4 + x) + tan² (π/4 − x) =
(a) a² + 1
(b) a² + 2
(c) a² − 2
(d) None of these

Answer 1

(c) $a^2-2$
$$\begin{gathered} \text{Given:} \\ \tan \left(\frac{\mathrm{\pi }}{4}+x\right)+\tan \left(\frac{\mathrm{\pi }}{4}-x\right)=a \\ \Rightarrow {\left[\tan \left(\frac{\mathrm{\pi }}{4}+x\right)+\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\right]}^2=a^2 \\ \Rightarrow {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)+2\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\tan \left(\frac{\mathrm{\pi }}{4}+x\right)=a^2 \\ \Rightarrow {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)=a^2-2\tan \left(\frac{\mathrm{\pi }}{4}-x\right)\tan \left(\frac{\mathrm{\pi }}{4}+x\right) \\ \Rightarrow {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)=a^2-2\left[\frac{\tan 45°-\tan x}{1+\tan 45°\tan x}\times \frac{\tan 45°+\tan x}{1-\tan 45°\tan x}\right] \\ \Rightarrow {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)=a^2-2\left[\frac{1°-\tan x}{1+\tan x}\times \frac{1+\tan x}{1-\tan x}\right] \\ \Rightarrow {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)=a^2-2\left(\frac{1-{\tan }^{2}x}{1-{\tan }^{2}x}\right) \\ \Rightarrow {\tan }^2\left(\frac{\mathrm{\pi }}{4}+x\right)+{\tan }^2\left(\frac{\mathrm{\pi }}{4}-x\right)=a^2-2 \end{gathered}$$