Question

Which of the following is the solution of set of the equation $2\cos^{- 1}(x) = \text{cot}^{- 1}\left( \dfrac{2x^{2} – 1}{2x\sqrt{1 – x^{2}}} \right)\$ ?
A. $(0,\ 1)$
B. $( - 1,\ 1) - \\{ 0\\}$
C. $( - 1,\ 0)$
D. $\lbrack – 1,\ 1\rbrack$

Answer 1

In this question, we need to find out the solution set of the values of $x$. First, let us substitute $\cos\ \theta$ in the place of $x$. Then we can solve it by considering the left and right part of the given expression separately . To solve the expression we have to use trigonometric formulas and identity. Then we have to equate both the left and right part of the expression. Using this we can easily find out the interval of $x$.

Complete step by step answer:
Given, $2\cos^{- 1}(x) = \cot^{- 1}\left( \dfrac{2x^{2} – 1}{2x\sqrt{1 – x^{2}}} \right)\$
Here we need to find out the interval of $x$. Let us substitute,
$x = \cos\ \theta$
$\Rightarrow \ x \in \lbrack – 1,\ 1\rbrack$
Since we know that the range of cosine function is $\lbrack – 1,\ 1\rbrack$
On substituting $x = \cos\ \theta$
We get,
$\Rightarrow \ 2\cos^{- 1}\left( \cos\ \theta \right) = \cot^{- 1}\left( \dfrac{2\left( \cos\ \theta \right)^{2} – 1}{2\left( \cos\ \theta \right)\sqrt{1 - \left( \cos\ \theta \right)^{2}}} \right)$

First let us consider the left part of the expression.
$\Rightarrow \ \ 2\cos^{- 1}\left( \cos\ \theta \right)$
We know that $\cos^{- 1}\left( \cos\ \theta \right) = \theta$
Hence we get,
$\Rightarrow \ 2(\theta)$
Thus $\theta \in \lbrack 0,\ \pi\rbrack$ •••(1)
Now let us consider the right part of the expression.
$\Rightarrow \ \cot^{- 1}\left( \dfrac{2\left( \cos\ \theta \right)^{2} – 1}{2\left( \cos\ \theta \right)\sqrt{1 - \left( \cos\ \theta \right)^{2}}} \right)$
We know that $\sin\ \theta = \sqrt{1 - \left( \cos\ \theta \right)^{2}}$

By substituting $\sin\ \theta$ in the place of $\sqrt{1 - \left( \cos\ \theta \right)^{2}}$ ,
We get,
$\Rightarrow \ \text{cot}^{- 1}\left( \dfrac{2\left( \cos\ \theta \right)^{2} – 1}{2\left( \cos\ \theta \right)\sin\ \theta} \right)$
We also know that $2\left( \cos\ \theta \right)^{2} – 1 = \cos\ 2\theta$ ,
Thus we get,
$\Rightarrow \ \text{cot}^{- 1}\left( \dfrac{\cos\ 2\theta}{2\left( {\cos\ \theta} \right)\sin\ \theta} \right)$
By using the formula $\sin\ 2\theta = 2\cos\ \theta\ \sin\ \theta$,
We get,
$\Rightarrow \ \text{cot}^{- 1}\left( \dfrac{\cos\ 2\theta}{\sin\ 2\theta} \right)$

We also know that cotangent is the ratio of the cosine function to the sine function.Thus we get,
$\Rightarrow \ \text{cot}^{- 1}\left( \cot\ 2\theta \right)$
We know that $\text{cot}^{- 1}\left( \cot\ 2\theta \right) = 2\theta$
By using this we get,
$\Rightarrow \ 2\theta$
$\Rightarrow \ 0 < 2\theta < \pi$
On dividing by $2$ ,
We get,
$0 < \theta < \dfrac{\pi}{2}$ •••(2)
Now on equating the expression (1) and (2) ,
We get
$0 < \theta < \dfrac{\pi}{2}$

By putting cosine function,
We get,
$\Rightarrow \ \cos\left( 0 \right) < \cos\left( \theta \right) < \cos\left( \dfrac{\pi}{2} \right)$
We know that $\cos(0)$ is $0$ and $\cos\left( \dfrac{\pi}{2} \right)$ is $1$
$\Rightarrow \ 0 < \cos\ \theta < 1$
By substituting $x = \cos\ \theta$ ,
We get,
$\Rightarrow \ 0 < x < 1$
Therefore $x \in (0,\ 1)$. Thus we get the solution of set of the equation $2\cos^{- 1}\left( x \right) = \cot^{- 1}\left( \dfrac{2x^{2} – 1}{2x\sqrt{1 – x^{2}}} \right)\$ is $\ (0,\ 1)$

Hence, the correct answer is option A.

Note: In order to solve these types of questions, we must have a stronger grip over the trigonometric properties . We must remember that the range of cosine function is $\lbrack - 1,1\rbrack$.The range of the cosine function helps in getting us the required interval of $x$ and helps us on the right track to reach our answer. We should also be careful while solving the inverse trigonometric functions.