Question

If 2.sin 2 theta =√3, then prove that 3 cot square theta -2 sin 0 =8.​

Answer 1

Question:

If 2•sin2@ = √3 , then prove that

3•(cot@)^2 - 2•sin@ = 8.

Given:

2•sin2@ = √3

To prove:

3•(cot@)^2 - 2•sin@ = 8

Proof:

We have ,

=> 2•sin2@ = √3

=> sin2@ = √3/2

=> sin2@ = sin60° { sin60° = √3/2 }

=> 2@ = 60°

=> @ = 60°/2

=> @ = 30°

Now, we have ;

=> LHS = 3•(cot@)^2 - 2•sin@

=> LHS = 3•(cot30°)^2 - 2•sin30°

{ @ = 30° }

=> LHS = 3•(√3)^2 - 2•(1/2)

{ cot30°= √3 , sin30°=1/2 }

=> LHS = 3•3 - 1

=> LHS = 9 - 1

=> LHS = 8

=> LHS = RHS

Hence proved.

Answer 2

Correct Question:

If 2sin 2 theta =√3, then prove that 3 cot square theta -2 sin 30 =8.

Step-by-step explanation:

For simplicity , we will use theta as alpha.

Now,

Given that,

$2 \sin(2 \alpha ) = \sqrt{3}$

$= > \sin(2 \alpha ) = \frac{ \sqrt{3} }{2} \\ \\ = > sin(2 \alpha ) = \sin( \frac{\pi}{ 3} ) \\ \\ = > 2 \alpha = \frac{\pi}{3} \\ \\ = > \alpha = \frac{\pi}{6}$

Now,

we know that,

$\cot( \frac{\pi}{6} ) = \sqrt{3}$

Therefore,

$= > { \cot }^{2} ( \frac{\pi}{3} ) = {( \sqrt{3}) }^{2} \\ \\ = > 3 { \cot }^{2} ( \frac{\pi}{6} ) = 3 \times 3 = 9 \\ \\ = > 3 { \cot}^{2}( \frac{\pi}{6} ) - 2 \sin(30) = 9-2×\frac{1}{2}$

$(°.° sin 30° = \frac{1}{2} )$

Hence,

$\bold{ 3 { \cot}^{2}( \frac{\pi}{6} ) - 2 \sin(30) = 8}$

Thus,

Proved