Question

If $\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$, then which among the following is true?
I. $\sin \alpha +\cos \alpha =\sqrt{2}$
II. $\sin \alpha +\cos \alpha =\sqrt{2}\sec \theta$
III. $\sin \alpha +\cos \alpha =\sqrt{2}\cos \theta$
IV. $\sin \alpha +\cos \alpha =\sqrt{2}\sin \theta$

• A. I is true
• B. II is true
• C. III is true
• D. IV is true

Answer 1

Answer: (C)

We have, $\tan \theta =\frac{\sin \alpha -\cos \alpha }{\sin \alpha +\cos \alpha }$
$\tan \theta =\frac{\tan \alpha -1}{\tan \alpha +1}\left(\begin{array}{l}\text{ dividing by }\cos \alpha \text{ in }\\ \text{ numerator and denominator }\end{array}\right)$
$=\frac{\tan \alpha -\tan \pi /4}{1+\tan \alpha \cdot \tan \frac{\pi }{4}}$
$=\tan \left(\alpha -\frac{\pi }{4}\right)$
$\Rightarrow \theta =\alpha -\frac{\pi }{4}$
$\Rightarrow \alpha =\left(\theta +\frac{\pi }{4}\right)$
$\therefore \sin \alpha +\cos \alpha =\sin \left(\theta +\frac{\pi }{4}\right)+\cos \left(\theta +\frac{\pi }{4}\right)$
$=\sin \theta \cdot \frac{1}{\sqrt{2}}+\cos \theta \cdot \frac{1}{\sqrt{2}}+\cos \theta \cdot \frac{1}{\sqrt{2}}-\sin \theta \cdot \frac{1}{\sqrt{2}}$
$=\frac{2\cos \theta }{\sqrt{2}}=\sqrt{2}\cos \theta$