If sinθ + sinΦ=a and cos θ +cosΦ=b prove that $tan\frac{\theta  – \phi }{2}$ =  $\pm \sqrt{\frac{4 – a^2 – b^2}{a^2 + b^2}}$

If sinθ + sinΦ=a and cos θ +cosΦ=b prove that $tan\frac{\theta  – \phi }{2}$ =  $\pm \sqrt{\frac{4 – a^2 – b^2}{a^2 + b^2}}$ =  $\pm \sqrt{\frac{4 – a^2 – b^2}{a^2 + b^2}}$

If sinθ + sinΦ=a and cos θ +cosΦ=b prove that $tan\frac{\theta  – \phi }{2}$ =  $\pm \sqrt{\frac{4 – a^2 – b^2}{a^2 + b^2}}$ =  $\pm \sqrt{\frac{4 – a^2 – b^2}{a^2 + b^2}}$

Solution:

sinθ + sinΦ=a —(i)

cos θ +cosΦ=b —-(ii)

Squaring equations (i) and (ii) and then adding we get,

(sinθ + sinΦ)²+( cos θ +cosΦ)²=a²+b²

or, sin²θ + 2 sinθ sinΦ + sin²Φ + cos²θ +2 cosθ cosΦ + cos²Φ = a²+b²

or, sin² θ+ cos² θ+ sin² θ+ sin² Φ+ 2 sinθ sinΦ+2 cosθ cosΦ= a²+b²

or, 1+1 +2(cosθ cosΦ+sinθ sinΦ)= a²+b²

or, 2+2 cos(θ-Φ)= a²+b²

or, 2cos(θ-Φ)= a²+b²-2

or,$\cos \left({\mathrm{\theta}} – {\mathrm{\phi}}\right)$ = $\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2 – 2}{2}$

$2c{\mathrm{os}}^2\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2} – 1$ = $\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2 – 2}{2}$

or,$2cos^2\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2 – 2}{2} + 1$

or,$2cos^2\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2}{2}$

or,$cos^2\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2}{4}$

or,$sec^2\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\frac{4}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$

or,$tan^2\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\frac{4}{{\mathrm{a}}^2 + {\mathrm{b}}^2} – 1$

or,$tan^2\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\frac{4 – a^2 – b^2}{a^2 + b^2}$

or,$tan\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\pm \sqrt{\frac{4 – a^2 – b^2}{a^2 + b^2}}$

$\therefore tan\frac{{\mathrm{\theta}} – {\mathrm{\phi}}}{2}$ = $\pm \sqrt{\frac{4 – a^2 – b^2}{a^2 + b^2}}$[Proved]