Trigonometric Function 

Prove the identity $\frac{\cos 2x}{\sin^{2} x + \frac{1}{2} \sin 2x } $ = $\cot x - 1.$

Trigonometry Functions
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1 Answer

Prove: \frac{\cos 2x}{\sin^{2} x + \frac{1}{2} \sin 2x} = \cot x - 1 

sin 
2
 x+ 
2
1
​
 sin2x
cos2x
​
 =cotx−1
 
Start with LHS: \frac{\cos 2x}{\sin^{2} x + \frac{1}{2} \sin 2x} 
sin 
2
 x+ 
2
1
​
 sin2x
cos2x
​
 
Use double angle identity for cosine: \cos 2x = \cos^2 x - \sin^2 xcos2x=cos 
2
 x−sin 
2
 x.
Factor the numerator: \frac{(\cos x + \sin x)(\cos x - \sin x)}{\sin^{2} x + \frac{1}{2} \sin 2x} 
sin 
2
 x+ 
2
1
​
 sin2x
(cosx+sinx)(cosx−sinx)
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