1.

Prove the left hand side of question ​

Answer»

LHS = \frac{tan^3\theta-1}{tan\theta-1}

APPLYING the formula a^3-b^3 = (a-b)(a^2+ab+b^2), we GET

LHS =\frac{(tan\theta-1)(tan^2\theta+tan\theta+1)}{(tan\theta-1)}

        = tan^2\theta+tan\theta+1

Applying the IDENTITY Tan^2\theta+1 = Sec^2\theta, we get

LHS = Sec^2\theta+Tan\theta = RHS

HOPE THIS HELPS!!


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