Question

Simplest Form :-Cot inv(sqrt( 1 + cos3x / 1 - cos3x) )

Answer 1

Please see the attachment. 

Answer 2

Answer:

$\cot^{-1}(\sqrt{\frac{1+\cos 3x}{1-\cos 3x}})=\frac{3x}{2}$

Step-by-step explanation:

Given : Expression $\cot^{-1}(\sqrt{\frac{1+\cos 3x}{1-\cos 3x}})$

To find : Simplify the expression ?

Solution :

We solve the expression by applying trigonometry properties,

$\cot^{-1}(\sqrt{\frac{1+\cos 3x}{1-\cos 3x}})$

Applying,

$1+\cos x=2\cos^2(\frac{x}{2})$

$1-\cos x=2\sin^2(\frac{x}{2})$

$=\cot^{-1}(\sqrt{\frac{2\cos^2(\frac{3x}{2})}{2\sin^2(\frac{3x}{2})}})$

$=\cot^{-1}(\sqrt{\frac{\cos^2(\frac{3x}{2})}{\sin^2(\frac{3x}{2})}})$

We know, $\cot x= \frac{\cos x}{\sin x}$

$=\cot^{-1}(\sqrt{\cot^2(\frac{3x}{2}})$

$=\cot^{-1}(cot(\frac{3x}{2})$

We know, $\cot^{-1}\cot (x)=x$

$=\frac{3x}{2}$

Therefore, $\cot^{-1}(\sqrt{\frac{1+\cos 3x}{1-\cos 3x}})=\frac{3x}{2}$