IIT JEE , AIEEE Forum - Mathematics , Trignometry

ramkumar_november

$tan\theta tan\alpha\;=\;\sqrt{\frac{a-b}{a+b}}$

$tan^2\alpha\;=\;\frac{(a-b)}{(a+b)tan^2\theta}$ .........(1)

now

$(a-bcos2\alpha)\;=\;\Bigg(a-b\bigg(\frac{1-tan^2\alpha}{1+tan^2\alpha}\bigg)\Bigg)$

= $\Bigg(\frac{a(1+tan^2\alpha)-b(1-tan^2\alpha)}{1+tan^2\alpha}\Bigg)$

substitute for $tan^2\alpha$ from (1) and then simplifying we get........

$(a-bcos2\alpha)\;=\;\frac{(a^2-b^2)(1+tan^2\theta)}{a(1+tan^2\theta)-b(1-tan^2\theta)}$

$(a-bcos2\alpha)\;=\;\frac{(a^2-b^2)}{a-b\bigg(\frac{1-tan^2\theta}{1+tan^2\theta}\bigg)}$

$(a-bcos2\alpha)\;=\;\frac{(a^2-b^2)}{a-bcos2\theta}$

$(a-bcos2\alpha)(a-bcos2\theta)\;=\;a^2-b^2$

which is independent of theta and alpha .......

please excuse me for not doing the simplification part as it is very tedious ........ i hope you can do it easily.........