Question

Prove the following expression such as $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$ .

Answer 1

The given expression has three parts. The first part, second part, and the third part of the expression are $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)$ , ${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ , and ${{\tan }^{2}}A$ respectively. Use the identity, $\cot A=\dfrac{1}{\tan A}$ and simplify the first part and second part of the expression. Now, solve it further and get the required result.

Complete step-by-step solution:
According to the question, we have to prove the given expression. We can observe that the given expression has three parts.
In the first part, we have
$\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)$ ………………………………….(1)
Similarly, in the second part, we have
${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$ …………………………………..(2)
Similarly, in the third part, we have
${{\tan }^{2}}A$ ………………………………………(3)
First of all, let us solve the first part, $\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)$
We know the identity that cotangent of an angle is reciprocal of the tangent of that angle, $\cot A=\dfrac{1}{\tan A}$ ………………………………………..(4)
Now, using equation (4) and on simplifying equation (1), we get

& =\left( \dfrac{1+{{\tan }^{2}}A}{1+\dfrac{1}{{{\tan }^{2}}A}} \right) \\\ & \Rightarrow \left( \dfrac{1+{{\tan }^{2}}A}{1+{{\tan }^{2}}A} \right){{\tan }^{2}}A \\\ \end{aligned}$$ $$\Rightarrow {{\tan }^{2}}A$$ ……………………………………….(5) We can observe that on solving the first part, we get $${{\tan }^{2}}A$$ . Now, from equation (3) and equation (5), we have $$\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\tan }^{2}}A$$ ……………………………………..(6) In the second part, we have $${{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}$$ . Using equation (4), the above equation can be simplified into simpler form. Now, from equation (2) and equation (4), we get $$\begin{aligned} & \Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}} \\\ & \Rightarrow {{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}} \\\ & \Rightarrow {{\left( \dfrac{1-\tan A}{\tan A-1} \right)}^{2}}{{\tan }^{2}}A \\\ & \Rightarrow {{\left( -1 \right)}^{2}}{{\tan }^{2}}A \\\ \end{aligned}$$ $$\Rightarrow {{\tan }^{2}}A$$ ………………………………..(7) Also, we can observe that on solving the second part, we get $${{\tan }^{2}}A$$ . Now, from equation (3) and equation (7), we get $$\Rightarrow {{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$$ ……………………………………………….(8) On comparing equation (6) and equation (8), we get $$\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$$ Hence, the given expression is proved. **Note:** For this type of question where we have two or three parts and we have to prove all of them equal to each other. Always approach this type of question by solving each part one by one and then, hold the equality between all the parts of the expression.