Question

The value of $\tan \theta \left( {1 - {{\cot }^2}\theta } \right)$ is equal to
(a). $\cot \theta \left( {1 - {{\tan }^2}\theta } \right)$
(b). $\cot \theta \left( {{{\tan }^2}\theta - 1} \right)$
(c). $\cot \theta {\tan ^2}\theta $
(d). $\tan \theta \cos e{c^2}\theta $

Answer 1

Hint:Here, we will use a trigonometric relation between $\tan \theta $ and $\cot \theta $ to solve the problem. This kind of problem is generally solved by seeing the relationship between the problem statement and the answer options to proceed step by step.

Complete step by step solution:
$\tan \theta \left( {1 - {{\cot }^2}\theta } \right)$ (This is the problem which we have to solve using trigonometric relation)

By opening the bracket, we will get:
= $\tan \theta - \tan \theta \times {\cot ^2}\theta ...........\left( 1 \right)$
(Let us consider this as equation 1)
Since, trigonometric relation between $\tan \theta $ and $\cot \theta $ is: $\cot = \frac{1}{{\tan \theta }} = \frac{1}{{\tan \theta }}$ and ${\cot ^2}\theta = \frac{1}{{{{\tan }^2}\theta }}$

By substituting the value ${\cot ^2}\theta $ , we can convert the equation (1) into:
= $\tan \theta - \tan \theta \times \frac{1}{{{{\tan }^2}\theta }}$
= $\tan \theta - \frac{1}{{\tan \theta }}$
We can see that now we only get the equation in $\tan \theta $

By solving further, we will get:
= $\frac{{{{\tan }^2}\theta - 1}}{{\tan \theta }}$
Now, by taking $\frac{1}{{\tan \theta }}$as common, we will get:
= $\frac{1}{{\tan \theta }}\left( {{{\tan }^2}\theta - 1} \right)$
Since, we know that $\frac{1}{{\tan \theta }} = \cot \theta $, we can convert it into:
= $\cot \theta \left( {{{\tan }^2}\theta - 1} \right)$

Hence, option (b) is the correct answer.

Note: In this kind of question, it is important to look at the answer options while looking towards the question and to proceed accordingly step by step. We can try to solve and match one of the options steps by step. For solving such a problem, we need to remember the trigonometric relations.