Question

prove that tan²(x)-sin²(x)=tan²(x)sin²(x)

Answer 1

Answer:

$\tan ^2\left(x\right)-\sin ^2\left(x\right)=\tan ^2\left(x\right)\sin ^2\left(x\right):\quad \mathrm{True}$

Step-by-step explanation:

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

Manipulating right side

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

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

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

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

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

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

$\mathrm{Express\:with\:sin,\:cos}$

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

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

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

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

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

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

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

$\left(\dfrac{\sin \left(x\right)}{\cos \left(x\right)}\right)^2$

$\mathrm{Apply\:exponent\:rule}:\quad \left(\dfrac{a}{b}\right)^c=\dfrac{a^c}{b^c}$

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

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

$\mathrm{Multiply\:fractions}:\quad \:a\cdot \dfrac{b}{c}=\dfrac{a\:\cdot \:b}{c}$

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

$\mathrm{Cancel\:the\:common\:factor:}\:\cos ^2\left(x\right)$

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

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

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

$\Rightarrow \mathrm{True}$