Question

Find an actual $\theta ,when\frac{cos\theta -sin\theta }{cos\theta +sin\theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

Answer 1

We have
$\frac{cos\theta -sin\theta }{cos\theta +sin\theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

By Applying Componendo and Dividendo Theorem

$\Rightarrow \frac{(cos\theta -sin\theta )+(cos\theta +sin\theta )}{(cos\theta -sin\theta )-(cos\theta +sin\theta )}=\frac{(1-\sqrt{3})+(1+\sqrt{3})}{(1-\sqrt{3})-(1+\sqrt{3})}$

$\Rightarrow \frac{2cos\theta }{-2sin\theta }=-\frac{2}{2\sqrt{3}}$
$\Rightarrow cot\theta =\frac{1}{\sqrt{3}}\Rightarrow tan\theta =\sqrt{3}\Rightarrow tan\theta =tan{60}^{\circ }\Rightarrow \theta ={60}^{\circ }$

$Alternate$

$\frac{cos\theta -sin\theta }{cos\theta +sin\theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$\Rightarrow \frac{\frac{cos\theta -sin\theta }{cos\theta }}{\frac{cos\theta +sin\theta }{cos\theta }}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

$\Rightarrow \frac{1-tan\theta }{1+tan\theta }=\frac{1-\sqrt{3}}{1+\sqrt{3}}$

By Comparing Both Sides

$\Rightarrow tan\theta =\sqrt{3}$

$\Rightarrow tan\theta =tan{60}^{\circ }$

$\Rightarrow \theta ={60}^{\circ }$