Question

$prove\:\tan ^2\left(x\right)-\sin ^2\left(x\right)=\tan ^2\left(x\right)\sin ^2\left(x\right)$

Answer 1

$\bf{GIVEN\:\::}$

$\bigstar\bf{Prove\:}:\bf{\tan ^2\left(x\right)-\sin ^2\left(x\right)=\tan ^2\left(x\right)\sin ^2\left(x\right)}$

$\bf{ANSWER\:\::}$

$\bf{Manipulating\:\:right\:\:side}$

$\tan ^2\left(x\right)\sin ^2\left(x\right)$

$\mathrm{Use\:the\:following\:identity}:\quad \sin ^2\left(x\right)=1-\cos ^2\left(x\right)$

$=\left(1-\cos ^2\left(x\right)\right)\tan ^2\left(x\right)$

$\mathrm{Expand}\:\left(1-\cos ^2\left(x\right)\right)\tan ^2\left(x\right)$

$\longrightarrow\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac$

$a=\tan ^2\left(x\right),\:b=1,\:c=\cos ^2\left(x\right)$

$=\tan ^2\left(x\right)\cdot \:1-\tan ^2\left(x\right)\cos ^2\left(x\right)$

$=1\cdot \tan ^2\left(x\right)-\tan ^2\left(x\right)\cos ^2\left(x\right)$

$=\tan ^2\left(x\right)-\cos ^2\left(x\right)\tan ^2\left(x\right)$

$-\cos ^2\left(x\right)\tan ^2\left(x\right)=-\sin ^2\left(x\right)$

$\mathrm{Using\:the\:Basic\:Trigonometric\:identity}:\quad \tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}$

$\tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}$

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

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

$\mathrm{Simplify}\:-\left(\frac{\sin \left(x\right)}{\cos \left(x\right)}\right)^2\cos ^2\left(x\right):\quad -\sin ^2\left(x\right)$

$\bigstar\bf{Result:\tan ^2\left(x\right)-\sin ^2\left(x\right)}$

$\longrightarrow\bf{R.H.S\:\:=\:\:L.H.S}$

$We\:showed\:that\:the\:two\:sides\:could\:take\:the\:same\:form$

$\bigstar\bf{Henced\:\: proved\:\:that} :\bf{\tan ^2\left(x\right)-\sin ^2\left(x\right)=\tan ^2\left(x\right)\sin ^2\left(x\right)}$