Question

If $\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$, then show that $\sin \alpha +\cos \alpha =\sqrt{2}\cos \theta$. [NCERT EXEMPLER]

Answer 1

$\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$

Dividing numerator and denominator on the RHS by $\cos \alpha$, we get

$$\begin{gathered} \tan \theta =\frac{{\displaystyle \frac{\sin \alpha }{\cos \alpha }}-1}{{\displaystyle \frac{\sin \alpha }{\cos \alpha }}+1} \\ \Rightarrow \tan \theta =\frac{\tan \alpha -\tan {\displaystyle \frac{\mathrm{\pi }}{4}}}{1+\tan \alpha \tan {\displaystyle \frac{\mathrm{\pi }}{4}}} \\ \Rightarrow \tan \theta =\tan \left(\alpha -\frac{\mathrm{\pi }}{4}\right) \\ \Rightarrow \theta =\alpha -\frac{\mathrm{\pi }}{4} \\ \mathrm{Or}\alpha =\frac{\mathrm{\pi }}{4}+\theta \end{gathered}$$

Now,

$$\begin{gathered} \sin \alpha +\cos \alpha \\ =\sin \left(\frac{\mathrm{\pi }}{4}+\theta \right)+\cos \left(\frac{\mathrm{\pi }}{4}+\theta \right) \\ =\sin \frac{\mathrm{\pi }}{4}\cos \theta +\cos \frac{\mathrm{\pi }}{4}\sin \theta +\cos \frac{\mathrm{\pi }}{4}\cos \theta -\sin \frac{\mathrm{\pi }}{4}\sin \theta \\ =\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{\sqrt{2}}\cos \theta -\frac{1}{\sqrt{2}}\sin \theta \\ =\frac{2}{\sqrt{2}}\cos \theta \\ =\sqrt{2}\cos \theta \end{gathered}$$

$\therefore \sin \alpha +\cos \alpha =\sqrt{2}\cos \theta$