Question

How to solve tan^-1(1/cos2x-sin2x/cos2x)

Answer 1

Answer:

$\boxed{\huge{ \dfrac{\pi}{4} - x}}$

Step-by-step explanation:

$\bold{To \: find}\longrightarrow \\ \: value \: of \\ {tan}^{ - 1} ( \: \dfrac{1}{cos2x} - \dfrac{sin2x}{cos2x} \: )$

$\bold{Concept \: used}\longrightarrow \\ \\1) sin \alpha = cos( \dfrac{\pi}{2} - \alpha ) \\2) cos \alpha = sin( \dfrac{\pi}{2} - \alpha ) \\ 3)1 - cos \alpha = 2 {sin}^{2} \dfrac{ \alpha }{2} \\ 4)sin2 \alpha = 2 \: sin \alpha \: cos \alpha \\ 4) \dfrac{sin \alpha }{cos \alpha } = tan \alpha$

$\bold{Solution}\longrightarrow \\ {tan}^{ - 1} ( \: \dfrac{1}{cos2x} - \dfrac{sin2x}{cos2x} \: )$

$= {tan}^{ - 1} \dfrac{1 - sin2 x}{cos2x}$

$= {tan}^{ - 1} \: \: ( \: \dfrac{1 - cos( \dfrac{\pi}{2} - 2x) }{sin( \dfrac{\pi}{2} - 2x)} \: )$

$= {tan}^{ - 1} \dfrac{2 {sin}^{2} \dfrac{1}{2} ( \dfrac{\pi}{2} - 2x) }{2 \: sin \dfrac{1}{2}( \dfrac{\pi}{2} -2 x) \: \: cos \dfrac{1}{2}( \dfrac{\pi}{2} - 2x) }$

${tan}^{ - 1} \dfrac{sin( \dfrac{\pi}{4} - x) }{cos( \dfrac{\pi}{4} - x) }$

$= {tan}^{ - 1} tan(\dfrac{\pi}{4} - x)$

$= \dfrac{\pi}{4} - x$