Question

Solve : ${\sin }^2\theta \tan \theta +{\cos }^2\theta \cot \theta -\sin 2\theta =1+\tan \theta +\cot \theta$

Answer 1

$si{n}^2\theta +co{s}^2\theta -si{n}^2\theta =1+tan\theta +cos\theta$

$\frac{si{n}^3\theta }{cos\theta }+\frac{co{s}^3\theta }{sin\theta }-2sin\theta cos\theta =1+\frac{sin\theta }{cos\theta }+\frac{cos\theta }{sin\theta }$

$\frac{si{n}^4\theta +co{s}^4\theta -2si{n}^2\theta co{s}^2\theta }{cos\theta sin\theta }$=$\frac{cos\theta sin\theta +si{n}^2\theta +co{s}^2\theta }{cos\theta sin\theta }$

$(si{n}^2\theta -co{s}^2\theta )=si{n}^2\theta +co{s}^2\theta +cos\theta sin\theta$

$co{s}^{2}2\theta =1+cos\theta sin\theta$

$co{s}^2\theta -\frac{sin2\theta }{2}=1$

$2co{s}^{2}2\theta -sin2\theta =2$