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prove sin theta-2sin cube theta/2cos cube theta-cos theta=tan theta

 let theta =x

so sin x - 2sin3x 

       2cos3x - cosx                                   = tanx

= sinx(1 - 2sin2x)      

      cosx(2cos2x - 1)

=sinx[1 -2(1 -cos2x)]

   cosx(2cos2x - 1)

=sinx[1 - 2 + 2cos2x]

   cosx(2cos2x - 1)

=sinx[2cos2x - 1]

     cosx(2cos2x - 1)

= sinx

     cosx

= tanx = rhs

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above answer is correct
 
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What is this method is poor
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Change the questions
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thanks for answering
 
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From where 1-2 came as common
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hi
sorry but  i'm in class 9
regards
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Sin 2 theta =2 sin theta cos theta
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SinQ-sin^3Q/cos^3Q-cosQ = tanQ

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Hi

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USE COS2A + SIN2A = 1 TO PROVE IT
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Wish it helps!!!

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I hope the below ans will help you

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solution for the given question is attached below! :)

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