IIT JEE , AIEEE Forum - Mathematics , Trignometry

ramkumar_november

let y = $tan^{-1}\Bigg(\frac{xcos\theta}{1-xsin\theta}\Bigg)-cot^{-1}\Bigg(\frac{cos\theta}{x-sin\theta}\Bigg)$

y = $tan^{-1}\Bigg(\frac{xcos\theta}{1-xsin\theta}\Bigg)-tan^{-1}\Bigg(\frac{x-sin\theta}{cos\theta}\Bigg)$

y = $tan^{-1}\Bigg(\frac{\frac{xcos\theta}{1-xsin\theta}-\frac{x-sin\theta}{cos\theta}}{1+\frac{x(x-sin\theta)}{1-xsin\theta}}\Bigg)$

y = $tan^{-1}\Bigg(\frac{xcos^2\theta-x+sin\theta+x^2sin\theta-xsin^2\theta}{cos\theta(1-xsin\theta+x^2-xsin\theta)}\Bigg)$

y = $tan^{-1}\Bigg(\frac{x(1-sin^2\theta)-x+sin\theta+x^2sin\theta-xsin^2\theta}{cos\theta(1-xsin\theta+x^2-xsin\theta)}\Bigg)$

y = $tan^{-1}\Bigg(\frac{-xsin^2\theta+sin\theta+x^2sin\theta-xsin^2\theta}{cos\theta(1-xsin\theta+x^2-xsin\theta)}\Bigg)$

y = $tan^{-1}\Bigg(\frac{sin\theta(1+x^2-2xsin\theta)}{cos\theta(1+x^2-2xsin\theta)}\Bigg)$

y = $tan^{-1}(tan\theta)$

y = $\theta$