Question

If tan (π/4 + θ) + tan (π/4 − θ) = a, then tan² (π/4 + θ) + tan² (π/4 − θ) =

(a) a² + 1
(b) a² + 2
(c) a² − 2
(d) None of these

Answer 1

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