Question

prove the trig identity: tan^2 x cos^2 x = (sec^2 x -1)(1-sin^4 x)/1+sin^2 x

Answer 1

Answer:

See steps below

Step-by-step explanation:

We need to work with each side of the equation at a time:

Left hand side:

Write all factors using the basic trig functions "sin" and "cos" exclusively:

$tan^2(x)\,cos^2(x)=\frac{sin^2(x)}{cos^2(x)} cos^2(x)=sin^2(x)$

now, let's work on the right side, having in mind the following identities:

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

b) $1-sin^4(x) =(1-sin^2(x))\,(1+sin^2(x))$

c) $1-sin^2(x)=cos^2(x)$

Then replacing we get:

$\frac{(sec^2(x)-1)\,(1-sin^4(x))}{1+sin^2(x)} =\frac{sin^2(x)(1+sin^2(x))(1-sin^2(x))}{cos^2(x)(1+sin^2(x)))} =\frac{sin^2(x)(1+sin^2(x))\,cos^2(x)}{cos^2(x)(1+sin^2(x)))}=sin^2(x)$

Therefore, we have proved that the two expressions are equal.