Question

If $sin\theta +cos\theta =0$ and $\theta$ lies in the fourth quadrant, find $sin\theta$ and $cos\theta$.

Answer 1

We have,

$sin\theta +cos\theta =0$

$\Rightarrow sin\theta =-cos\theta .....\left(i\right)$

$\Rightarrow \frac{sin\theta }{cos\theta }=-1$

We know that,

$se{c}^2\theta -ta{n}^2\theta =1$

$\Rightarrow sec\theta =1+ta{n}^2\theta$

$\Rightarrow sec\theta =\pm \sqrt{1+ta{n}^2\theta }$

In the 4th quadrant sec $\theta$ positive.

$\therefore sec\theta =\sqrt{1+ta{n}^2\theta }$

$=\sqrt{1+(-1{)}^2}$

$=\sqrt{1+1}=\sqrt{2}$

$\therefore cos\theta =\frac{1}{sec\theta }=\frac{1}{\sqrt{2}}$

Putting $cos\theta =\frac{1}{\sqrt{2}}$ in equation (i),we get,

$sin\theta =-\left(\frac{1}{\sqrt{2}}\right)=-\frac{1}{\sqrt{2}}$

Hence, $sin\theta =-\frac{1}{\sqrt{2}}$ and $cos\theta =\frac{1}{\sqrt{2}}.$