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}}$
$=\frac{2 \cos \theta}{\sqrt{2}}=\sqrt{2} \cos \theta$