Question

if sec theta =2 ,then find : (a) cot theta + cosec theta ​

Answer 1

Given :-

$\sf sec \theta = 2$

To Find :-

$\sf cot \theta + cosec \theta$

Solution :-

We know :-

$\sf cos\theta = \dfrac{1}{sec \theta }$

$\sf\therefore cos \theta = \dfrac{1}{2}$

We know :-

$\bf sin^2 \theta +cos^2 \theta= 1$

$\longrightarrow \sf sin^2 \theta + \left( \dfrac{1}{2} \right)^2= 1 \\\\\longrightarrow sin^2 \theta +\dfrac{1}{4} = 1 \\\\\longrightarrow sin^2 \theta = 1 - \dfrac{1}{4} \\\\\longrightarrow sin^2 \theta = \dfrac{4-1}{4} \\\\\longrightarrow sin^2 \theta = \dfrac{3}{4}\\\\\longrightarrow sin \theta = \sqrt{\dfrac{3}{4}} \\\\\longrightarrow sin \theta = \dfrac{\sqrt{3} }{2}$

$\bullet \bf \ cosec \theta = \dfrac{1}{sin \theta }$

$\longrightarrow \sf cosec \theta = \dfrac{2}{\sqrt{3} }$

$\bullet \bf \ cot \theta = \dfrac{cos \theta }{sin \theta }$

$\longrightarrow \sf cot\theta = \dfrac{1}{2} \times \dfrac{2}{\sqrt{3} }\\\\\longrightarrow cot \theta = \dfrac{1}{\sqrt{3} }$

$\sf cot \theta + cosec \theta = \dfrac{1}{\sqrt{3}} +\dfrac{2}{\sqrt{3}}$

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