if a cosφ=b cosθ then show that atanθ+btanφ=(a+b) tan$\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}$

if a cosφ=b cosθ then show that atanθ+btanφ=(a+b) tan$\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}$

if a cosφ=b cosθ then show that atanθ+btanφ=(a+b) tan$\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}$

$acos\phi$ = $bcos\theta$

$\Rightarrow \frac{{\mathrm{a}}}{\cos {\mathrm{\theta}}} = \frac{{\mathrm{b}}}{\cos {\mathrm{\phi}}}$

${\mathrm{Let}},\frac{{\mathrm{a}}}{\cos {\mathrm{\theta}}}$ = $\frac{{\mathrm{b}}}{\cos {\mathrm{\phi}}}$ = K[say]

$\therefore {\mathrm{a}} = {\mathrm{K}}cos{\mathrm{\theta}}{\mathrm{and}}{\mathrm{b}} = {\mathrm{K}}cos{\mathrm{\phi}}$

$\therefore {\mathrm{LHS}}$

= ${\mathrm{a}}tan{\mathrm{\theta}} + {\mathrm{b}}tan{\mathrm{\phi}}$

= ${\mathrm{K}}cos{\mathrm{\theta}}tan{\mathrm{\theta}} + {\mathrm{K}}cos{\mathrm{\phi}}tan{\mathrm{\phi}}$

= ${\mathrm{K}}cos{\mathrm{\theta}} \times \frac{\sin {\mathrm{\theta}}}{\cos {\mathrm{\theta}}} + {\mathrm{K}}cos{\mathrm{\phi}} \times \frac{\sin {\mathrm{\phi}}}{\cos {\mathrm{\phi}}}$

= ${\mathrm{K}}sin{\mathrm{\theta}} + {\mathrm{K}}sin{\mathrm{\phi}}$

= ${\mathrm{K}}(sin{\mathrm{\theta}} + sin{\mathrm{\phi}})$

= ${\mathrm{K}} \times 2sin\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}cos\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$

= ${\mathrm{K}} \times 2cos\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}cos\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2} \times \frac{sin\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}}{\cos \frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}}$

= ${\mathrm{K}}(cos{\mathrm{\theta}} + cos{\mathrm{\phi}})tan\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}$

= $({\mathrm{K}}cos{\mathrm{\theta}} + {\mathrm{K}}cos{\mathrm{\phi}})tan\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}$

= $({\mathrm{a}} + {\mathrm{b}})tan\frac{{\mathrm{\theta}} + {\mathrm{\phi}}}{2}$

= ${\mathrm{RHS}}({\mathrm{\Pr oved}})$