Question

Prove that $\frac{\tan \theta -\cot \theta }{\sin \theta \cos \theta }={\tan }^2\theta -{\cot }^2\theta$.

Answer 1

$\frac{tan\theta -cot\theta }{sin\theta cos\theta }=ta{n}^2\theta -co{t}^2\theta$

solving LHS

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

$\Rightarrow se{c}^2\theta -cose{c}^2\theta$

We know $1+ta{n}^2\theta =se{c}^2\theta$ & $1+co{t}^2\theta =co{t}^2\theta$

$\Rightarrow 1+ta{n}^2\theta -1-co{t}^2\theta$

$\Rightarrow ta{n}^2\theta -co{t}^2\theta =RHS$