Math, asked by harshmali10101, 6 months ago

tan5A-ran3A/tan5A+tan3A=sin2A/sin8A

Answers

Answered by sritarutvik

0

Step-by-step explanation:

tan5A-tan3A/tan5A+tan3A

=(sin5A/cos5A - sin3A/cos3A ) /(sin5A/cos5A + sin3A/cos3A)

=((sin5Acos3A - sin3Acos5A)/cos5Acos3A) / ((

sin5Acos3A + sin3Acos5A)/cos5Acos3A)

=sin(5A-3A)/sn(5A+3A)

=sin2A/sin8A

Answered by vankaniriddhi

0

Answer:

yes put tanA =sinA/cosA

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