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

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)))

fromlefthandsidesprovethatsinαsinβcosα+cosβ=2tanα2tanβ21tan2α2tan2β2

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

LHS    a=tan(α/2)   b=tan(β/2)  (((2a×2b)/((1+a^2 )(1+b^2 )))/(((1−a^2 )/(1+a^2 ))+((1−b^2 )/(1+b^2 ))))  ((4ab)/(1+b^2 −a^2 −a^2 b^2 +1−b^2 +a^2 −a^2 b^2 ))  =((4ab)/(2(1−a^2 b^2 )))=((2ab)/(1−a^2 b^2 ))

LHSa=tanα2b=tanβ22a×2b(1+a2)(1+b2)1a21+a2+1b21+b24ab1+b2a2a2b2+1b2+a2a2b2=4ab2(1a2b2)=2ab1a2b2

Commented by ajfour last updated on 24/Dec/18

very nice Sir!

veryniceSir!

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

thank you sir...

thankyousir...

Commented by peter frank last updated on 24/Dec/18

thank you

thankyou

Answered by peter frank last updated on 25/Dec/18

((sinαsin β)/(cos α+cos β))=((4sin (α/2)sin (α/2).sin (β/2)cos (β/2))/(cosα+sin β))  cos α=cos^2  (α/2)−sin^2 (α/2)  cos β=cos^2  (β/2)−sin^2 (β/2)  ((4sin (α/2)sin (α/2).sin (β/2)cos (β/2))/((cos^2 (α/2)−sin^2 (α/2))+(cos^2 (β/2)−sin^2 (β/2))))  divide by cos^2 (α/2)cos^2 (β/2) both  numerator and denominator  ((4tan (α/2)tan (β/2))/(sec^2 (β/2)(1−tan^2  (α/2))+sec^2 (β/2)(1−tan^2 (β/2))))    ((4tan (α/2)tan (β/2))/(2−2tan^2 (α/2)tan^2 (β/2)))  ((2tan (α/2)tan (β/2))/(1−tan^2 (α/2)tan^2 (β/2)))

sinαsinβcosα+cosβ=4sinα2sinα2.sinβ2cosβ2cosα+sinβcosα=cos2α2sin2α2cosβ=cos2β2sin2β24sinα2sinα2.sinβ2cosβ2(cos2α2sin2α2)+(cos2β2sin2β2)dividebycos2α2cos2β2bothnumeratoranddenominator4tanα2tanβ2sec2β2(1tan2α2)+sec2β2(1tan2β2)4tanα2tanβ222tan2α2tan2β22tanα2tanβ21tan2α2tan2β2

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