Respuesta :
[tex]\frac{cot^2x -cosec\ x}{cosec^2x} = cos^2x - sin\ x[/tex]
Solution:
Given that we have to prove:
[tex]\frac{cot^2x -cosec\ x}{cosec^2x} = cos^2x - sin\ x[/tex]
Take the L.H.S of above
[tex]\frac{cot^2x -cosec\ x}{cosec^2x}[/tex]
Simplify
[tex]\frac{cot^2x}{cosec^2x} -\frac{cosec\ x}{cosec^2x}\\\\Simplify\\\\\frac{cot^2x}{cosec^2x} -\frac{1}{cosec\ x}[/tex]
We know that,
[tex]sin\ x = \frac{1}{cosec\ x}[/tex]
Therefore,
[tex]\frac{cot^2x}{cosec^2x} -sin\ x[/tex]
Also we know that,
[tex]cot^2x = \frac{cos^2\ x}{sin^2\ x}[/tex]
[tex]cosec^2\ x = \frac{1}{sin^2\ x}[/tex]
Thus we get,
[tex]\frac{\frac{cos^2\ x}{sin^2 x}}{\frac{1}{sin^2\ x}} - sin\ x\\\\Simplify\\\\\frac{cos^2\ x}{sin^2 x} \times sin^2\ x - sin x\\\\Simplify\\\\cos^2\ x - sin\ x[/tex]
Thus,
L.H.S = R.H.S
Thus proved
