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tanA = a tanB, sinA = bsinB rArr (b^(2) ...

`tanA = a tanB, sinA = bsinB rArr (b^(2) - 1)/(a^(2) - 1)` =

A

`sin^(2)A`

B

`sin^(3)A`

C

`cos^(2)A`

D

`cos6(3)A`

Text Solution

Verified by Experts

The correct Answer is:
C
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