Question

If Cosec A = 2, find the value of 1/tan+ sin/1+cos

Answer 1

Given :-

$\sf cosecA = 2$

To Find :-

The value of :-

$\sf \dfrac{1}{tanA} + \dfrac{sinA}{1+cosA}$

Solution :-

We know :-

$\sf sinA = \dfrac{1}{cosecA}$

$\therefore \sf sinA = \dfrac{1}{2}$

We know :-

$\sf cos^2A = 1-sin^2A$

$\longrightarrow \sf cos^2A = 1- \left( \dfrac{1}{2} \right)^2 \\\\\longrightarrow cos^2A = 1- \dfrac{1}{4} \\\\\longrightarrow cos^2A = \dfrac{3}{4} \\\\\longrightarrow cosA = \dfrac{\sqrt{3}}{2}$

We know :-

$\sf tanA = \dfrac{sinA}{cosA}$

$\therefore \sf tanA = \dfrac{1}{2} \times \dfrac{2}{\sqrt{3}}$

           $= \sf \dfrac{1}{\sqrt{3}}$

$\longrightarrow \sf \dfrac{1}{tanA}+\dfrac{sinA}{1+cosA} = \dfrac{1}{\dfrac{1}{\sqrt{3}}} + \dfrac{\dfrac{1}{2}}{1+ \dfrac{\sqrt{3}}{2}} \\\\\longrightarrow \dfrac{1}{tanA} +\dfrac{sinA}{1+cosA} = \sqrt{3} + \dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}} \\\\\longrightarrow \dfrac{1}{tanA} +\dfrac{sinA}{1+cosA} = \sqrt{3} +\dfrac{1}{2+\sqrt{3}} \\\\\longrightarrow \dfrac{1}{tanA} +\dfrac{sinA}{1+cosA} = \dfrac{2\sqrt{3} +3+1}{2+\sqrt{3}}$

$\longrightarrow \sf \dfrac{1}{tanA} +\dfrac{sinA}{1+cosA} = \dfrac{2\sqrt{3}+4} {2+\sqrt{3}} \\\\\longrightarrow \dfrac{1}{tanA}+\dfrac{sinA}{1+cosA} = \dfrac{2(\sqrt{3}+2)}{2+\sqrt{3}} \\\\\longrightarrow \bf \dfrac{1}{tanA} +\dfrac{sinA}{1+cosA} = 2$