I am unable to solve this question. Plz give the solution for this...the correct answer option (C).
Anushka Khuspe , 6 Years ago
Grade
Trigonometry> I am unable to solve this question. Plz g...
1 Answers
Aditya Gupta
Last Activity: 6 Years ago
thank u anu for askin.
2sin^2b=1-cos2b
also, 4cos(a+b)sinasinb=2cos(a+b)[cos(a-b)-cos(a+b)]=cos2a+cos2b-1-cos2(a+b) [here we have recursively applied standard product formulae]
so, adding we get
2sin^2b+ 4cos(a+b)sinasinb+cos2(a+b)
=cos2a
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