Question

Identity transformations of Trigonometric Expressions.
prove the following identities.
$\frac{sin(\pi -\alpha )}{sin\alpha -cos\alpha tan\frac{\alpha }{2}}+cos(\pi -\alpha )=1.$

Answer 1

$\frac{sin(\pi -\alpha )}{sin\alpha -cos\alpha tan\alpha /2}+cos(\pi -\alpha )=1$

$LHS=\frac{sin(\pi -\alpha )}{sin2-cos2tan\alpha /2}+cos(\pi -\alpha )$

$\therefore cos(\pi -\alpha )=-cos\alpha$ (in $I{I}^{nd}$ quadrant cos is -ve)

$\therefore sin(\pi -\alpha )=sin\alpha$ in $I{I}^{nd}$ quadrant sin is +ve

and $tan\frac{\alpha }{2}=\frac{sin\alpha }{1+cos\alpha }$ {By properties of trigonometric fraction}.

So

$LHS=\frac{sin\alpha }{sin\alpha -\frac{cos\alpha .sin\alpha }{1+cos\alpha }}-cos\alpha$

$LHS=\frac{sin\alpha }{sin\alpha (1-\frac{cos\alpha }{1+cos\alpha })}-cos\alpha$

$LHS=\frac{1+cos\alpha }{1+cos\alpha -cos\alpha }-cos\alpha =1+cos\alpha -coa\alpha$

$=1=RHS$ Proved.