Question

Prove that :
$\frac{tan\theta -cot\theta }{sin\theta cos\theta }=se{c}^2\theta -cose{c}^2\theta =ta{n}^2\theta -co{t}^2\theta$

Answer 1

We have,
LHS = $\frac{tan\theta -cot\theta }{sin\theta cos\theta }=\frac{\frac{sin\theta }{cos\theta }-\frac{cos\theta }{sin\theta }}{sin\theta cos\theta }=\frac{\frac{si{n}^2\theta -co{s}^2\theta }{sin\theta cos\theta }}{cos\theta sin\theta }$

$\Rightarrow$ LHS = $\frac{si{n}^2\theta -co{s}^2\theta }{si{n}^2\theta co{s}^2\theta }$

$\Rightarrow$ LHS = $\frac{si{n}^2\theta }{si{n}^2\theta co{s}^2\theta }-\frac{co{s}^2\theta }{si{n}^2\theta co{s}^2\theta }$

$\Rightarrow$ LHS = $\frac{1}{co{s}^2\theta }-\frac{1}{si{n}^2\theta }$

$\Rightarrow$ LHS = $se{c}^2\theta -cose{c}^2\theta =(1+ta{n}^2\theta )-(1+co{t}^2\theta )=ta{n}^2\theta -co{t}^2\theta =RHS.$