Question

prove that 1+ tan^2 theta /1+cot^2 theta=[1- tan theta / 1- cot theta ]^2

Answer 1

Answer:

Concept:

Proving Trigonometric Identities or Equations.

Step-by-step explanation:

Given:

$\frac{1+\tan ^{2} \Theta}{1+\cot ^{2} \Theta}=\left(\frac{1-\tan \Theta}{1-\cot \Theta}\right)^{2}=\tan ^{2} \Theta .$

Find:

Prove that  $\frac{1+\tan ^{2} \Theta}{1+\cot ^{2} \Theta}=\left(\frac{1-\tan \Theta}{1-\cot \Theta}\right)^{2}=\tan ^{2} \Theta .$

Solution:

                            $\begin{aligned}&\frac{1+\tan ^{2} \theta}{1+\cot ^{2} \theta}=\frac{1+\tan ^{2} \theta}{1+\frac{1}{\tan ^{2} \theta}} \\\\&\left(\frac{1-\tan \theta}{1-\cot \theta}\right)^{2}=\frac{1+\tan ^{2} \theta}{\left(\frac{\tan ^{2} \theta+1}{\tan ^{2} \theta}\right)} \\\\&=\left(\frac{1-\tan \theta}{1-\frac{1}{\tan \theta}}\right)^{2}=\tan ^{2} \theta \\\\&=\left(\tan \theta\left(\frac{1-\tan \theta}{\tan \theta-1}\right)\right)^{2} \\\\&=(-\tan \theta)^{2} \\\\&=\tan ^2 \theta\end{aligned}$

Hence proved.

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