Question

1 + 10 square theta divided by 1 + cot square theta is equal to 1 minus 10 theta divided by 1 minus cot theta whole square

Answer 1

Step-by-step explanation:

The given equation is:

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

Taking the LHS of the above equation, we have

\frac{1+tan^2{\theta}}{1+cot^2{theta}}

=\frac{sec^2\theta}{cosec^2\theta}

=\frac{sin^2\theta}{cos^2\theta}

=tan^2{\theta}

Now, taking the RHS of the above equation, we have

(\frac{1-tan{\theta}}{1-cot{\theta}})^{2}

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

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

=tan^2{\theta}

Hence, LHS=RHS, thus proved.

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