# Math Help - trig fix

1. ## trig fix

[(1)/(sec^(2)T-cos^(2)T)+(1)/cosec^(2)T-sin^(2)T)]sin^(2)Tcos^(2)T=(1-cos^(2)Tsin^(2)T)/(2+cos^(2)Tsin^(2))

2. ## Re: trig fix

Hello, srirahulan!

A few more parentheses would helped.
As it is, I had to guess what you meant . . .

$\left[\frac{1}{\sec^2\!x - \cos^2\!x} + \frac{1}{\csc^2\!x - \sin^2\!x}\right]\sin^2\!x\cos^2\!x \;=\;\frac{1-\sin^2\!x\cos^2\!x}{2 + \sin^2\!x\cos^2\!x}$

The left side is: . $\left[\frac{1}{\frac{1}{\cos^2\!x} - \cos^2\!x} + \frac{1}{\frac{1}{\sin^2\!x} - \sin^2x}}\right]\sin^2\!x\cos^2\!x$

. . $=\;\left[\frac{\cos^2\!x}{1 - \cos^4\!x} + \frac{\sin^2\!x}{1 - \sin^4\!x}\right]\sin^2\!x\cos^2\!x$

. . $=\;\left[\frac{\cos^2\!x}{(1-\cos^2\!x)(1+\cos^2\!x)} + \frac{\sin^2\!x}{(1-\sin^2\!x)(1+\sin^2\!x)}\right]\sin^2\!x\cos^2\!x$

. . $=\;\left[\frac{\cos^2\!x}{\sin^2\!x(1+\cos^2\!x)} + \frac{\sin^2\!x}{\cos^2\!x(1 + \sin^2\!x)}\right]\sin^2\!x\cos^2\!x$

. . $=\;\left[\frac{\cos^4\!x(1+\sin^2\!x) + \sin^4\!x(1+\cos^2\!x)}{\sin^2\!x\cos^2\!x(1+\sin^ 2\!x)(1+\cos^2\!x)}\right]\sin^2\!x\cos^2\!x$

. . $=\;\frac{\cos^4\!x + \sin^2\!x\cos^4\!x + \sin^4\!x + \sin^4\!x\cos^2\!x}{(1+\cos^2\!x)(1 + \sin^2\!x)}$

. . $=\;\frac{\cos^4\!x+\sin^4\!x + \sin^2\!x\cos^2\!x\overbrace{(\cos^2\!x+\sin^2\!x) }^{\text{This is 1}}}{(1 + \cos^2\!x)(1+\sin^2\!x)} \;=\;\frac{\cos^4\!x + \sin^4\!x + \sin^2\!x\cos^2\!x}{(1+\cos^2\!x)(1+\sin^2\!x)}$

. . $=\;\frac{\cos^4\!x + 2\cos^2\!x\sin^2\!x + \sin^4\!x - \sin^2\!x\cos^2\!x}{(1+\cos^2\!x)(1+\sin^2\!x)}$

. . $=\;\frac{\overbrace{(\cos^2\!x + \sin^2\!x)^2}^{\text{This is }1^2} - \sin^2\!x\cos^2\!x}{(1+\cos^2x)(1+\sin^2\!x)} \;=\;\frac{1 - \sin^2\!x\cos^2\!x}{(1+\cos^2\!x)(1+\sin^2\!x)}$

. . $=\;\frac{1-\sin^2\!x\cos^2\!x}{1 + \underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} + \sin^2\!x\cos^2\!x} \;=\;\frac{1-\sin^2\!x\cos^2\!x}{2 + \sin^2\!x\cos^2\!x}$