Question

12. Prove that 2 cosec4x - sec2x =1-tanx/1+tanx cosec 2x​

Answer 1

✪PROOF✪

$\:\:\:\:\: LHS$

$= 2csc(4x) - sec(2x)$

$= \frac{2}{sin(4x)} - \frac{1}{cos(2x)}$

$= \frac{2}{2sin(2x)cos(2x)} - \frac{1}{cos(2x)}$

$= \frac{1 - sin(2x)}{sin(2x)cos(2x)}$

$= \frac{1 - 2sin(x)cos(x)}{cos(2x)}csc(2x)$

$= \frac{sin^2x + cos^2x- 2sin(x) cos(x)}{cos^2x-sin^2x}csc(2x)$

$= \frac{[cos(x)- sin(x)]^2}{[cos(x) -sin(x)][cos(x)+sin(x)]}csc(2x)$

$= \frac{cos(x)- sin(x)}{cos(x)+sin(x)}csc(2x)$

$= \frac{1- \frac{sin(x)}{cos(x)}}{1+\frac{sin(x)}{cos(x)}}csc(2x)$

$= \frac{1- tan(x)}{1+tan(x)}csc(2x)$

$= RHS$

Hence Proved