Verify: tan^2x - sin^2x= (tan^2x)(sin^2x)

Saturday, March 19, 2011

We know the $\displaystyle{{\sin}}^{{2}}+{{\cos}}^{{x}}={1}$ .

Therefore $\displaystyle{{\sin}}^{{2}}{x}={1}-{{\cos}}^{{2}}{x}$ .

$\displaystyle{\tan{{x}}}=\frac{{\sin{{x}}}}{{\cos{{x}}}}.$ We use these identities in the course of the solution.

RHS: $\displaystyle{{\tan}}^{{2}}{x}\cdot{{\sin}}^{{2}}{x}={{\tan}}^{{2}}{x}{\left({1}-{{\cos}}^{{2}}{x}\right)}$ ,

$\displaystyle={{\tan}}^{{2}}{x}-{\left({{\tan}}^{{2}}{x}\right)}\cdot{{\cos{{x}}}}^{{2}}$ .

$\displaystyle={{\tan}}^{{2}}{x}-{{\left(\frac{{\sin{{x}}}}{{\cos{{x}}}}\right)}}^{{2}}\cdot{{\cos}}^{{2}}{x}.$

$\displaystyle={{\tan}}^{{2}}{x}-{\left({{\sin}}^{{2}}\frac{{x}}{{{\cos}}^{{2}}}{x}\right)}{{\cos}}^{{2}}{x}$

$\displaystyle={{\tan}}^{{2}}{x}-{{\sin}}^{{2}}{x}\cdot{{\cos}}^{{2}}\frac{{x}}{{{\cos}}^{{2}}}{x}$

$\displaystyle={{\tan}}^{{2}}{x}-{{\sin}}^{{2}}{x}$ which is LHS.

Notes