from left hand sides prove that ((sinαsin β)/(cos α+cos β))=((2tan(α/2) tan (β/2))/(1−tan^2 (α/2)tan^2 (β/2)))


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Question Number 51150 by peter frank last updated on 24/Dec/18

$f r o m l e f t h a n d s i d e s$ $p r o v e t h a t$ $\frac{s i n α \sin β}{\cos α + \cos β} = \frac{2 \tan \frac{α}{2} \tan \frac{β}{2}}{1 - \tan^{2} \frac{α}{2} \tan^{2} \frac{β}{2}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

	$L H S$ $a = t a n \frac{α}{2} b = t a n \frac{β}{2}$ $\frac{\frac{2 a \times 2 b}{(1 + a^{2}) (1 + b^{2})}}{\frac{1 - a^{2}}{1 + a^{2}} + \frac{1 - b^{2}}{1 + b^{2}}}$ $\frac{4 a b}{1 + b^{2} - a^{2} - a^{2} b^{2} + 1 - b^{2} + a^{2} - a^{2} b^{2}}$ $= \frac{4 a b}{2 (1 - a^{2} b^{2})} = \frac{2 a b}{1 - a^{2} b^{2}}$
	Commented by ajfour last updated on 24/Dec/18

	$v e r y n i c e S i r!$
	Commented by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

	$t h a n k y o u s i r . . .$
	Commented by peter frank last updated on 24/Dec/18

	$t h a n k y o u$
	Answered by peter frank last updated on 25/Dec/18

	$\frac{s i n α \sin β}{\cos α + \cos β} = \frac{4 \sin \frac{α}{2} \sin \frac{α}{2} . \sin \frac{β}{2} \cos \frac{β}{2}}{c o s α + \sin β}$ $\cos α = \cos^{2} \frac{α}{2} - \sin^{2} \frac{α}{2}$ $\cos β = \cos^{2} \frac{β}{2} - \sin^{2} \frac{β}{2}$ $\frac{4 \sin \frac{α}{2} \sin \frac{α}{2} . \sin \frac{β}{2} \cos \frac{β}{2}}{(\cos^{2} \frac{α}{2} - \sin^{2} \frac{α}{2}) + (\cos^{2} \frac{β}{2} - \sin^{2} \frac{β}{2})}$ $d i v i d e b y \cos^{2} \frac{α}{2} \cos^{2} \frac{β}{2} b o t h n u m e r a t o r a n d d e n o m i n a t o r$ $\frac{4 \tan \frac{α}{2} \tan \frac{β}{2}}{\sec^{2} \frac{β}{2} (1 - \tan^{2} \frac{α}{2}) + \sec^{2} \frac{β}{2} (1 - \tan^{2} \frac{β}{2})}$ $\frac{4 \tan \frac{α}{2} \tan \frac{β}{2}}{2 - 2 \tan^{2} \frac{α}{2} \tan^{2} \frac{β}{2}}$ $\frac{2 \tan \frac{α}{2} \tan \frac{β}{2}}{1 - \tan^{2} \frac{α}{2} \tan^{2} \frac{β}{2}}$
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