Question

If $\tan \left( {{\cos }^{-1}}x \right)=\sin \left( {{\cot }^{-1}}\dfrac{1}{2} \right)$ then x is equal to –
(a). $\dfrac{1}{\sqrt{5}}$
(b). $\dfrac{2}{\sqrt{5}}$
(c). $\dfrac{3}{\sqrt{5}}$
(d). $\dfrac{\sqrt{5}}{3}$

Answer 1

Hint: Try to use the identity ${{\cos }^{-1}}x={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ and then write $\tan \left( {{\cos }^{-1}}x \right)$ as $\tan \left( {{\tan }^{-1}}\dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ or $\dfrac{\sqrt{1-{{x}^{2}}}}{x}$. After that use the identity ${{\cot }^{-1}}x={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)$ and the write $\sin \left( {{\cot }^{-1}}x \right)$ as $\sin \left( {{\sin }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)$ or $\dfrac{1}{\sqrt{1+{{x}^{2}}}}$. Hence do according to the question.

Complete step-by-step answer:

In the question we are given that the equation, $\tan \left( {{\cos }^{-1}}x \right)$ = $\sin \left( {{\cot }^{-1}}\dfrac{1}{2} \right)$ and we have to find the value of x such that it satisfies.
Before proceeding let us know what are inverse trigonometric functions, inverse trigonometric functions of trigonometric functions. Specifically they are inverse of sine, cosine, tangent, cotangent, secant, cosecant functions, and are used to obtain an angle from any of the angle’s trigonometric ratios.
There are certain notations which are used. Some of the most common notation is using arc sin (x), arc cos (x), arc tan (x) instead of ${{\sin }^{-1}}\left( x \right),{{\cos }^{-1}}\left( x \right)$ and ${{\tan }^{-1}}\left( x \right)$. These arises from geometric relationships. When measuring in radians, an angle $\theta$ radians will correspond to an arc whose length is $r\theta$, where r is radius of circle. Thus in the unit circle, “the arc whose cosine is x” is same as “ the angle whose cosine is x”, because length of arc of circle in radii is same as the measurement of angle in radius.
We are given that $\tan \left( {{\cos }^{-1}}x \right)$ is equal to $\sin \left( {{\cot }^{-1}}\left( \dfrac{1}{2} \right) \right)$. So here we will use the identity of inverse trigonometric function which is ${{\cos }^{-1}}x={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$. So, $\tan \left( {{\cos }^{-1}}x \right)$ is equal to $\tan \left( {{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right) \right)$ or $\dfrac{\sqrt{1-{{x}^{2}}}}{x}$.
Also there is an identity that ${{\cot }^{-1}}x$ which is also equal to ${{\sin }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$. So, $\sin \left( {{\cot }^{-1}}x \right)$ is equal to $\sin \left( {{\sin }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}} \right)$ or $\dfrac{1}{\sqrt{1+{{x}^{2}}}}$.
In the question $\sin \left( {{\cot }^{-1}}\left( \dfrac{1}{2} \right) \right)$ was there so it can be written as $\dfrac{1}{\sqrt{1+{{\left( \dfrac{1}{2} \right)}^{2}}}}$ or $\dfrac{1}{\sqrt{1+\dfrac{1}{4}}}$.
Hence the value of $\sin \left( {{\cot }^{-1}}\left( \dfrac{1}{2} \right) \right)$ is $\dfrac{1}{\sqrt{\dfrac{5}{4}}}$ or $\dfrac{2}{\sqrt{5}}$.
We know that $\tan \left( {{\cos }^{-1}}x \right)$ can be written as $\tan \left( {{\tan }^{-1}}\dfrac{\sqrt{1-{{x}^{2}}}}{x} \right)$ or $\dfrac{\sqrt{1-{{x}^{2}}}}{x}$.
So now according to the question we can say that the value of $\dfrac{\sqrt{1-{{x}^{2}}}}{x}$ is equal to $\dfrac{2}{\sqrt{5}}$.
So, we can write that,
\Rightarrow $$$$\dfrac{\sqrt{1-{{x}^{2}}}}{x} = $\dfrac{2}{\sqrt{5}}$
We will square both the sides of the equation so we get,
\Rightarrow $$$$\dfrac{1-{{x}^{2}}}{{{x}^{2}}}=\dfrac{4}{5}
Or we can write it as,
\Rightarrow $$$$\dfrac{1}{{{x}^{2}}}-1=\dfrac{4}{5}
Now adding 1 to both the sides so we get,
$\Rightarrow \dfrac{1}{{{x}^{2}}}-1+1=\dfrac{4}{5}+1$
Or, $\dfrac{1}{{{x}^{2}}}=\dfrac{9}{5}$
So the value of ${{x}^{2}}=\dfrac{5}{9}$ or $x=\pm \dfrac{\sqrt{5}}{3}$.
Hence the correct option is (d).

Note: The identities used in the question such as ${{\cos }^{-1}}x={{\tan }^{-1}}\dfrac{\sqrt{1-{{x}^{2}}}}{x}$ and ${{\cot }^{-1}}x={{\sin }^{-1}}\dfrac{1}{\sqrt{1+{{x}^{2}}}}$ are formed by considering a right angled triangle with two sides 1 and x and hypotenuse be $\sqrt{1+{{x}^{2}}}$.