Question

if $\sqrt{2}tan2\theta =\sqrt{6}and{0}^{\circ }<2\theta <{90}^{\circ },findthevalueofsin\theta +\sqrt{3}cos\theta -2{tan}^2\theta$

Answer 1

Given, $\sqrt{2}tan2\theta =\sqrt{6}$
$tan2\theta =\sqrt{6}/\sqrt{2}$
=$\sqrt{3}$
= $tan{60}^o$
$⟹2\theta ={60}^o$
$\theta ={30}^o$
Now,
$sin\theta +\sqrt{3}cos\theta –2ta{n}^2\theta$
= $sin{30}^o+\sqrt{3}cos{30}^o–2ta{n}^2{30}^o$
= $1/2+\sqrt{3}\times \sqrt{3}/2–2(1/\sqrt{3}{)}^2$
=$1/2+3/2–2/3=4/2–2/3$
= $(12-4)/6$
= $8/6$
=$4/3$