Question

prove that sinA+sin2A+sin3A÷cosA+cos2A+cos3A =tan 2A
​

Answer 1

We have to prove that,

(sinA + sin2A + sin3A)/(cosA + cos2A + cos3A) = tan2A

proof : LHS = (sinA + sin2A + sin3A)/(cosA + cos2A + cos3A)

= {(sinA + sin3A) + sin2A}/{cosA + cos3A) + cos2A}

we know, sin x + sin y = 2sin(x + y)/2 cos(x - y)/2

cos x + cos y = 2cos(x + y)/2 cos(x - y)/2

= {2sin(A + 3A)/2 cos(3A - A)/2 + sin2A}/{2cos(A + 3A)/2 cos(3A - A)/2 + cos2A}

= {2sin2A cosA + sin2A}/{2cos2A cosA + cos2A}

= {sin2A(2cosA + 1)}/{cos2A(2cosA + 1)}

= sin2A/cos2A = tan2A = RHS

also read similar questions : prove that sina+cosa/sina-cosa+sina-cosa/sina+cosa=2/sin2a-cos2a =2 sec^2a/tan^2a-1

https://brainly.in/question/27674742

prove that sin3A/sinA + cos3A/cosA = 4cos2A

https://brainly.in/question/16120497