Question

Prove sin a + sin2a + sin 3a / cos a + cos 2a +cos 3a = tan 2 a

Answer 1

Answer:

ur answer

Step-by-step explanation:

(sinA+sin3A)/(cosA+cos3A)

= (sinA+3sinA-4sin^3A)/(cosA+4cos^3A-3cosA)

= (4sinA-4sin^3A)/(4cos^3A-2cosA)

= [4sinA(1-sin^2A)]/[2cosA(2cos^2A-1)]

= (4sinAcos^2A)/(2cosAcos2A)

= (2sinAcosA)/cos2A

= sin2A/cos2A

= tan2A

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Answer 2

Answer:

Step-by-step explanation:

sin a + sin2a + sin 3a / cos a + cos 2a +cos 3a

Numerator  N = sin2a + sin 3a +sina

=sin2a + 2sin(3a+a)/2cos(3a-a)/2

=sin2a+2sin2a cosa

N=sin2a(1+2cosa)

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Denominator  D =cosa+cos3a+cos2a

=2cos(3a+a)2 cos(3a-a)/2 +cos 2a

=2cos2acosa + cos2a

D=cos2a(1+2cosa)

Thus result=N/D sin2a(1+2cosa) / cos2a(1+2cosa)

=sin 2a/cos 2a

=tan 2a=RHS