Question

if 2 sin 2 theta =√3 then prove that 3 cot square theta -2sin theta = 8​

Answer 1

Answer:

8

Step-by-step explanation:

$2 \sin(2 \alpha ) = \sqrt{3} \\ \sin(2 \alpha) = \frac{ \sqrt{3} }{2} \\ 2\alpha = { \sin }^{ - 1} \sin( \frac{\pi}{3} ) \\ \alpha = \frac{\pi}{6} \\now \\ 3 { \cot( \alpha ) }^{2} - 2 \sin( \alpha ) \\ put \: \alpha = \frac{\pi}{6} \\ 3 { \cot( \frac{\pi}{6} ) }^{2} - 2 \sin( \frac{\pi}{6} ) \\ 9 - 2 \times \frac{1}{2} \\ 9 - 1 \\ 8$

Formula used

Sin(π/6) = sqrt3/2

cot(π/6) = sqrt3

Answer 2

Answer with Step-by-step explanation:

We are given that

$2sin2 \theta=\sqrt3$

We have to prove that

$3cot^2\theta-2sin\theta=8$

$sin2\theta=\frac{\sqrt3}{2}$

$2\theta=sin^{-1}\frac{\sqrt3}{2}$

$2\theta=sin^{-1}(sin\frac{\pi}{3}})$

($sin\frac{\pi}{3}=\frac{\sqrt3}{2}$)

$2\theta=\frac{\pi}{3}$

$\theta=\frac{\pi}{6}$

$3cot^2\theta-2sin\theta=3cot^2(\frac{\pi}{6})-2sin(\frac{\pi}{6})$

$3cot^2\theta-2sin\theta=3\times 3-2\times \frac{1}{2}=9-1=8$

Because $cot\frac{\pi}{6}=\sqrt3, sin\frac{\pi}{6}=\frac{1}{2}$)

Hence, proved