Question

simplify the following expressions.
$\frac{1+cos\alpha }{1-cos\alpha }ta{n}^2\frac{\alpha }{2}-co{s}^2\alpha .$

Answer 1

$\frac{1+cos\alpha }{1-cos\alpha }ta{n}^2\frac{\alpha }{2}-co{s}^2\alpha$

We have formula $\frac{1-cos\theta }{1+cos\theta }=ta{n}^2\frac{\theta }{2}$

$\Rightarrow \frac{1+cos\theta }{1-cos\theta }=co{t}^2\frac{\theta }{2}$

$\Rightarrow \frac{1+cos\alpha }{1-cos\alpha }ta{n}^2\frac{\alpha }{2}-co{s}^2\alpha$

$\Rightarrow co{t}^2\frac{\alpha }{2}.ta{n}^2\frac{\alpha }{2}-co{s}^2\alpha$ $[\because ta{n}^2\alpha =\frac{1}{co{t}^2\alpha }]$

$\Rightarrow co{t}^2\frac{\alpha }{2}\times \frac{1}{co{t}^2\frac{\alpha }{2}}-co{s}^2\alpha$

$\Rightarrow 1-co{s}^2\alpha$ $[\because co{s}^2\alpha +si{n}^2\alpha =1]$

$\Rightarrow si{n}^2\alpha$