Question

please answer it correctly​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\dfrac{sin(\theta)-cos(\theta)}{sin(\theta)+cos(\theta)}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}}$

Using componendo and dividendo,

$\sf{\implies\,\dfrac{\{sin(\theta)-cos(\theta)\}+\{sin(\theta)+cos(\theta)\}}{\{sin(\theta)-cos(\theta)\}-\{sin(\theta)+cos(\theta)\}}=\dfrac{(\sqrt{3}-1)+(\sqrt{3}+1)}{(\sqrt{3}-1)-(\sqrt{3}+1)}}$

$\sf{\implies\,\dfrac{2sin(\theta)}{-2cos(\theta)}=\dfrac{2\sqrt{3}}{-2}}$

$\sf{\implies\,\dfrac{sin(\theta)}{cos(\theta)}=\sqrt{3}}$

$\sf{\implies\,tan(\theta)=\sqrt{3}}$

$\sf{\implies\,\theta=\dfrac{\pi}{3}}$

$\sf{\implies\,\dfrac{\theta}{2}=\dfrac{\pi}{6}}$

Now,

$\sf{\dfrac{sin\bigg(\dfrac{\theta}{2}\bigg)-cos\bigg(\dfrac{\theta}{2}\bigg)}{tan\bigg(\dfrac{\theta}{2}\bigg)+cot\bigg(\dfrac{\theta}{2}\bigg)}}$

$\sf{=\dfrac{sin\bigg(\dfrac{\pi}{6}\bigg)-cos\bigg(\dfrac{\pi}{6}\bigg)}{tan\bigg(\dfrac{\pi}{6}\bigg)+cot\bigg(\dfrac{\pi}{6}\bigg)}}$

$\sf{=\dfrac{\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}}{\dfrac{1}{\sqrt{3}}+\sqrt{3}}}$

$\sf{=\dfrac{\dfrac{1-\sqrt{3}}{2}}{\dfrac{1+3}{\sqrt{3}}}}$

$\sf{=\dfrac{\dfrac{1-\sqrt{3}}{2}}{\dfrac{4}{\sqrt{3}}}}$

$\sf{=\dfrac{(1-\sqrt{3})\sqrt{3}}{8}}$

$\sf{=\dfrac{\sqrt{3}-3}{8}}$