Question

If ${tan}^2\alpha =1+2{tan}^2\beta$ then prove that ${\cos }^2\beta =1+2co{s}^2\alpha$

Answer 1

We have, ${tan}^2\alpha =1+2{tan}^2\beta$

$\frac{{\sin }^2\alpha }{{\cos }^2\alpha }=1+2\frac{{\sin }^2\beta }{{\cos }^2\beta }$

$\Rightarrow {\sin }^2\alpha {\cos }^2\beta ={\cos }^2\alpha ({\cos }^2\beta +{\sin }^2\beta +{\sin }^2\beta )$

$\Rightarrow {\sin }^2\alpha {\cos }^2\beta ={\cos }^2\alpha (1+{\sin }^2\beta )$

$\Rightarrow {\sin }^2\alpha (1-{\sin }^2\beta )=(1-{\sin }^2\alpha )(1+{\sin }^2\beta )$

$\Rightarrow {\sin }^2\alpha -{\sin }^2\alpha {\sin }^2\beta =1-{\sin }^2\alpha +{\sin }^2\beta -{\sin }^2\alpha {\sin }^2\beta$

$\Rightarrow 2{\sin }^2\alpha =1+{\sin }^2\beta \Rightarrow co{s}^2\alpha +{\sin }^2\beta =0$

$\Rightarrow {\cos }^2\beta =1+2{\cos }^2\alpha$

Hence proved