Question

If $tan\theta$ = $\frac{sin\alpha -cos\alpha }{sin\alpha +cos\alpha }$, then $sin\alpha +cos\alpha$ and

$sin\alpha -cos\alpha$ must be equal to

• A. $\sqrt{2}cos\theta ,\sqrt{2}sin\theta$
• B. $\sqrt{2}sin\theta ,\sqrt{2}cos\theta$
• C. $\sqrt{2}sin\theta ,\sqrt{2}sin\theta$
• D. $\sqrt{2}cos\theta ,\sqrt{2}cos\theta$

Answer 1

The correct option is A

$\sqrt{2}cos\theta ,\sqrt{2}sin\theta$

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

$\Rightarrow$ $tan\theta$ = $\frac{sin(\alpha -\frac{\pi }{4})}{cos(\alpha -\frac{\pi }{4})}$ $\Rightarrow$ $tan\theta$ = $tan(\alpha -\frac{\pi }{4})$

$\Rightarrow$ $\theta$ = $\alpha$ - $\frac{\pi }{4}$ $\Rightarrow$ $\alpha$ = $\theta$ + $\frac{\pi }{4}$

Hence, $sin\alpha +cos\alpha$ = $sin(\theta +\frac{\pi }{4})+cos(\theta +\frac{\pi }{4})$

= $\sqrt{2}cos\theta$

And $sin\alpha -cos\alpha$ = $sin(\theta +\frac{\pi }{4})$ - $cos(\theta +\frac{\pi }{4})$

= $\frac{1}{\sqrt{2}}$ $sin\theta$ + $\frac{1}{\sqrt{2}}$ $cos\theta$ - $\frac{1}{\sqrt{2}}$ $cos\theta$ + $\frac{1}{\sqrt{2}}$ $sin\theta$

= $\frac{2}{\sqrt{2}}$ $sin\theta$ = $\sqrt{2}sin\theta =\sqrt{2}sin\theta$.