Question

Prove that : $1+tan\alpha .tan\left(\frac{\alpha }{2}\right)=sec\alpha$

Answer 1

Prove $\to$ tan a tan $\frac{a}{2}+1=secx$

$LHS=(tana\times tan\frac{a}{2})+1$

$=\frac{2tan\left(\frac{a}{2}\right)}{1-ta{n}^2\left(\frac{a}{2}\right)}\left(tan\frac{a}{2}\right)+1$

$=\frac{2ta{n}^2\frac{a}{2}+1-ta{n}^2\frac{a}{2}}{1-ta{n}^2\frac{a}{2}}$

$=\frac{1+ta{n}^2\left(\frac{a}{2}\right)}{1-ta{n}^2\left(\frac{a}{2}\right)}$

Now we know that $cosx=\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}$

so $\frac{1}{cosx}=secx=RHS$

Proved