Question

Find the value of $\cos [{\tan ^{ - 1}}(\sin ({\cot ^{ - 1}}x))]$
A.$\sqrt {\dfrac{{{x^2} + 2}}{{{x^2} + 3}}}$
B.$\sqrt {\dfrac{{{x^2} + 2}}{{{x^2} + 1}}}$
C.$\sqrt {\dfrac{{{x^2} + 1}}{{{x^2} + 2}}}$
D.$\sqrt {\dfrac{{{x^2} - 1}}{{{x^2} + 1}}}$

Answer 1

To solve such questions, where there are many trigonometric functions one within another, it is best to eliminate the innermost functions by changing them with the help of either some trigonometric property or by substituting. Here too we will use both the property as well as substitution of the functions into such a way that they can be eliminated.

Complete step-by-step answer:
The given expression is $\cos [{\tan ^{ - 1}}(\sin ({\cot ^{ - 1}}x))]$ .
We first begin with the innermost function which is inverse cotangent.
As we know that ${\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$ , we can also write this as ${\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x$.
So, the overall expression becomes $\cos [{\tan ^{ - 1}}(\sin (\dfrac{\pi }{2} - {\tan ^{ - 1}}x))]$ .
As we know $\sin (\dfrac{\pi }{2} - {\tan ^{ - 1}}x) = \cos ({\tan ^{ - 1}}x)$ because sine and cosine are complementary functions.
The expression is now written as $\cos [{\tan ^{ - 1}}(\cos ({\tan ^{ - 1}}x))]$.
For solving further, we have to convert the angle of cosine into inverse cosine function so that they cancel out each other. For this we assume ${\tan ^{ - 1}}x = \theta$, so $\tan \theta = x$ .
Now, by the definition of tangent function, $\tan \theta = \dfrac{{Opposite}}{{Adjacent}} = \dfrac{x}{1}$ .
So the hypotenuse is given by $\sqrt {{x^2} + 1}$. Now by definition of cosine function, $\cos \theta = \dfrac{{Adjacent}}{{Hypotenuse}} = \dfrac{1}{{\sqrt {{x^2} + 1} }}$ .
Thus $\theta$ can be written as $\theta = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {{x^2} + 1} }}$ by taking the inverse cosine function on both sides.
Now, we can substitute the value of ${\tan ^{ - 1}}x = \theta = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {{x^2} + 1} }}$ in the expression.
This gives $\cos [{\tan ^{ - 1}}(\cos ({\cos ^{ - 1}}\dfrac{1}{{\sqrt {{x^2} + 1} }}))]$ . Here cosine and cosine inverse function cancel each other giving $\cos [{\tan ^{ - 1}}(\dfrac{1}{{\sqrt {{x^2} + 1} }})]$.
We now have to repeat the above steps and get cosine inverse function instead of tangent inverse.
To ease the calculations, we first substitute $\dfrac{1}{{\sqrt {{x^2} + 1} }} = \alpha$, so the expression is now written as $\cos [{\tan ^{ - 1}}(\alpha )]$.
Let us assume that ${\tan ^{ - 1}}\alpha = \varphi$, which on simplifying gives $\tan \varphi = \alpha$.
Now, by definition of tangent function, $\tan \varphi = \dfrac{{Opposite}}{{Adjacent}} = \dfrac{\alpha }{1}$ .
So the hypotenuse is given by $\sqrt {{\alpha ^2} + 1}$ . Now by definition of cosine function, $\cos \varphi = \dfrac{{Adjacent}}{{Hypotenuse}} = \dfrac{1}{{\sqrt {{\alpha ^2} + 1} }}$ .
Thus $\varphi$ can be written as $\varphi = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {{\alpha ^2} + 1} }}$ by taking the inverse cosine function on both sides.
Now, we can substitute the value of ${\tan ^{ - 1}}\alpha = \varphi = {\cos ^{ - 1}}\dfrac{1}{{\sqrt {{\alpha ^2} + 1} }}$ in the expression.
This gives $\cos ({\cos ^{ - 1}}\dfrac{1}{{\sqrt {{\alpha ^2} + 1} }})$ . Here cosine and cosine inverse function cancel each other giving $\dfrac{1}{{\sqrt {{\alpha ^2} + 1} }}$ .
Now, we can substitute the value of $\alpha$ .
That gives $\dfrac{1}{{\sqrt {{{(\dfrac{1}{{\sqrt {{x^2} + 1} }})}^2} + 1} }}$ .
This can now be simplified as

$$\dfrac{1}{{\sqrt {{{(\dfrac{1}{{\sqrt {{x^2} + 1} }})}^2} + 1} }} \\\ = \dfrac{1}{{\sqrt {\dfrac{1}{{{x^2} + 1}} + 1} }} \\\ = \dfrac{1}{{\sqrt {\dfrac{{1 + {x^2} + 1}}{{{x^2} + 1}}} }} \\\ = \sqrt {\dfrac{{{x^2} + 1}}{{{x^2} + 2}}} \\\$$

Hence, the correct option is option (C).

Note: While solving such types of questions involving lots of trigonometric functions, it is best to draw a rough triangle with the angles marked in it to avoid confusion between different angles. Also, one must solve these questions in a step-by-step manner eliminating each function one-by-one.