If sin theta + sin phi = a and cos theta + co | vedclass

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(B) Given: $\sin 	heta + \sin \phi = a$ $(i)$ and $\cos 	heta + \cos \phi = b$ $(ii)$.Squaring and adding $(i)$ and $(ii)$:$(\sin 	heta + \sin \phi

Vedclass · Accepted Answer

$\sqrt{\frac{4 - a^2 - b^2}{a^2 + b^2}}$

Vedclass · Answer

(B) Given: $\sin 	heta + \sin \phi = a$ $(i)$ and $\cos 	heta + \cos \phi = b$ $(ii)$.Squaring and adding $(i)$ and $(ii)$:$(\sin 	heta + \sin \phi)^2 + (\cos 	heta + \cos \phi)^2 = a^2 + b^2$$(\sin^2 	heta + \cos^2 	heta) + (\sin^2 \phi + \cos^2 \phi) + 2(\sin 	heta \sin \phi + \cos 	heta \cos \phi) = a^2 + b^2$$1 + 1 + 2 \cos(	heta - \phi) = a^2 + b^2$$2 \cos(	heta - \phi) = a^2 + b^2 - 2$$\cos(	heta - \phi) = \frac{a^2 + b^2 - 2}{2}$.Using the identity $\cos x = \frac{1 - 	an^2(x/2)}{1 + 	an^2(x/2)}$:$\frac{1 - 	an^2(\frac{	heta - \phi}{2})}{1 + 	an^2(\frac{	heta - \phi}{2})} = \frac{a^2 + b^2 - 2}{2}$.Applying componendo and dividendo:$\frac{(1 + 	an^2(\frac{	heta - \phi}{2})) + (1 - 	an^2(\frac{	heta - \phi}{2}))}{(1 + 	an^2(\frac{	heta - \phi}{2})) - (1 - 	an^2(\frac{	heta - \phi}{2}))} = \frac{2 + (a^2 + b^2 - 2)}{2 - (a^2 + b^2 - 2)}$$\frac{2}{2 	an^2(\frac{	heta - \phi}{2})} = \frac{a^2 + b^2}{4 - a^2 - b^2}$$	an^2(\frac{	heta - \phi}{2}) = \frac{4 - a^2 - b^2}{a^2 + b^2}$$	an(\frac{	heta - \phi}{2}) = \sqrt{\frac{4 - a^2 - b^2}{a^2 + b^2}}$.

If $\sin \theta + \sin \phi = a$ and $\cos \theta + \cos \phi = b,$ then $\tan \frac{\theta - \phi}{2}$ is equal to

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