Question

sin^3(π+x) sec^2(π-x)tan(2π-x)/cos^2(π/2+x)sin(π-x)cosec^2-x=tan^x​

Answer 1

$\textbf{To prove:}$

$\dfrac{sin^3(\pi+x)\,sec^2(\pi-x)\,tan(2\pi-x)}{cos^2(\frac{\pi}{2}+x)\,sin(\pi-x)\,cosec^2(-x)}=tan^3x$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{sin^3(\pi+x)\,sec^2(\pi-x)\,tan(2\pi-x)}{cos^2(\frac{\pi}{2}+x)\,sin(\pi-x)\,cosec^2(-x)}$

$=\dfrac{(sin(\pi+x))^3\,(sec(\pi-x))^2\,tan(2\pi-x)}{(cos(\frac{\pi}{2}+x))^2\,sin(\pi-x)\,(cosec(-x))^2}$

$=\dfrac{(-sin\,x)^3\,(-sec\,x)^2\,(-tan\,x)}{(-sin\,x)^2\,(sin\,x)\,(-cosec\,x)^2}$

$=\dfrac{(-sin^3x)\,sec^2x\,(-tan\,x)}{sin^2x\,(sin\,x)\,cosec^2x}$

$=\dfrac{(-sin^3x)\,sec^2x\,(-tan\,x)}{sin^2x\,(sin\,x)\,cosec^2x}$

$=\dfrac{sin^3x\,(\frac{1}{cos^2x})\,tan\,x}{sin^2x\,(sin\,x)\,(\frac{1}{sin^2x})}$

$=\dfrac{sin^3x\,(\frac{1}{cos^2x})\,tan\,x}{sin\,x}$

$=sin^2x\,(\frac{1}{cos^2x})\,tan\,x$

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

$=tan^2x\,tan\,x$

$=tan^3x$

$\implies\boxed{\bf\dfrac{sin^3(\pi+x)\,sec^2(\pi-x)\,tan(2\pi-x)}{cos^2(\frac{\pi}{2}+x)\,sin(\pi-x)\,cosec^2(-x)}=tan^3x}$

Find more:

Tan(π/4+x)-tan(π/4-x)=2tan 2x

https://brainly.in/question/10999329

Answer 2

Answer:

it will be helpful !

for more answers like and follow