Question

Prove that tan inverse 3 sin 2 alpha upon 5 + 3 cos 2 alpha + tan inverse 1 by 4 10 alpha equals to alpha

Answer 1

Answer:

Step-by-step explanation:

It's given that,

$tan^{-1} \frac{3sin2\alpha }{5+3 cos 2\alpha }+tan^{-1} (\frac{1}{4}tan\alpha ) = \alpha$

solution:

LHS = $tan^{-1} \frac{3sin2\alpha }{5+3 cos 2\alpha } + tan^{-1} (\frac{1}{4}tan\alpha )$

$= tan^{-1} \frac{6 tan\alpha }{8+2tan^{2}\alpha } + tan^{-1} (\frac{1}{4}tan\alpha )$

$= tan^{-1} \frac{3 tan \alpha }{4+tan^{2}\alpha } + tan^{-1} (\frac{1}{4}tan\alpha )$

$= tan^{-1} (\frac{\frac{3tan\alpha }{4+tan^{2} \alpha}+ \frac{1}{4} tan\alpha}{1 -\frac{3 tan^2 \alpha }{16+4 tan^2\alpha }})$

=$tan^-1{ {\frac{(16+tan^2\alpha)tan\alpha }{16+tan^2\alpha } }$ }

= $tan^-1 (tan\alpha )$

=$\alpha$

= RHS