Math, asked by saketh2642, 10 months ago

Cos2a+2cos4a+cos6a/cosa+2cos3a+cos5a=cosa-sina*tan3a

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Answered by Anonymous

1

Answer:

Cos2a+2cos4a+cos6a/cosa+2cos3a+cos5a=cosa-sina*tan3aa

Answered by adirajgawale07

2

Answer:

Step-by-step explanation:

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