Question

If \[\tan ({\cos ^{ - 1}}x) = \sin ({\cot ^{ - 1}}\dfrac{1}{2})\], then find the value of \[x\].
A) \[x = - \dfrac{{\sqrt 5 }}{3}\]
B) \[x = + \dfrac{{\sqrt 5 }}{3}\]
C) \[x = - \dfrac{2}{3}\]
D) \[x = + \dfrac{2}{3}\]

Answer 1

Hint: In the given question, we have to apply trigonometric identities to solve the question. We will have to find the hypotenuse of the triangle to arrive at the value of \[\sin \theta \] and then use trigonometric ratios to solve \[\tan \theta \].

Complete step by step solution:

Let us solve the equation as follows:

First let us consider RHS term \[\sin ({\cot ^{ - 1}}\dfrac{1}{2})\]

Using the trigonometric identity \[{\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\dfrac{1}{\theta }\], we get,

 \[ = \sin ({\tan ^{ - 1}}2)\]

Using the formula, \[\tan \theta = \dfrac{{Opposite\,side\,}}{{Adjacent\,side\,}}\], we can draw following diagram: ($\theta$ is the angle ACB.)

Since \[{\tan ^{ - 1}} = 2\], we get

\[AB = 2\]and

\[BC = 1\]

Using Pythagoras Theorem (which states that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse), we get,

\[AC = \sqrt {A{B^2} + B{C^2}} \]

\[AC = \sqrt {{{(2)}^2} + {1^2}} \]

\[\Rightarrow AC = \sqrt {4 + 1} \]

\[\Rightarrow AC = \sqrt 5 \]

Therefore, we can get value of \[\sin \theta \],

\[\sin \theta = \dfrac{{AB}}{{AC}} = \dfrac{2}{{\sqrt 5 }}\]

Hence RHS= \[\dfrac{2}{{\sqrt 5 }}\]

Now we can proceed to LHS side of equation \[\tan ({\cos ^{ - 1}}x)\]:

Using the THS result, we will get,

\[\tan ({\cos ^{ - 1}}x) = \dfrac{2}{{\sqrt 5 }}\]

We can write the equation as follows:

\[{\cos ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{2}{{\sqrt 5 }}\]

Let \[{\tan ^{ - 1}}\dfrac{2}{{\sqrt 5 }} = t\]

Therefore, we can say that,

\[\dfrac{2}{{\sqrt 5 }} = \tan t\]

Now let us draw the diagram again as follows:

 

Here \[\tan t = \dfrac{{A'B'}}{{B'C'}} = \dfrac{2}{{\sqrt 5 }}\]

Using Pythagoras Theorem, we get,

\[A'C' = \sqrt {A'B{'^2} + B'C{'^2}} \]

\[A'C' = \sqrt {{{(2)}^2} + {{(\sqrt 5 )}^2}} \]

\[A'C' = \sqrt {4 + 5} \]

\[A'C' = \sqrt 9 \]

\[A'C' = 3\]

Therefore, we can get the value of \[\cos t = \dfrac{{B'C'}}{{A'C'}}\]

\[\cos t = \dfrac{{\sqrt 5 }}{3}\]

Since \[{\tan ^{ - 1}}\dfrac{2}{{\sqrt 5 }} = t\], we get,

\[{\cos ^{ - 1}}x = t\]

Using the property \[{\cos ^{ - 1}}A = B\]so \[A = \cos B\], we will get:

\[x = \cos t\]

\[x = \dfrac{{\sqrt 5 }}{3}\]

Hence, option (B) \[x = + \dfrac{{\sqrt 5 }}{3}\] is the correct answer.

Note:

1) \[{\cot ^{ - 1}}\theta = {\tan ^{ - 1}}\dfrac{1}{\theta }\] is proved as below:

Let \[{\cot ^{ - 1}}(x) = \theta \]

So, we get \[x = \cot \theta \]

We know that cotangent is the inverse of tangent. So, we get,

\[\dfrac{1}{x} = \tan \theta \]

Therefore, we can say that,

\[\theta = {\tan ^{ - 1}}(\dfrac{1}{x}) = {\cot ^{ - 1}}(x)\]

2) Meaning of the terms used to find the missing value in diagram are clarified below:

Sine: The ratio of side opposite to given angle and hypotenuse is called sine. It is denoted as \[\sin \theta \].

\[\sin \theta = \dfrac{{Side\,opposite\,to\,given\,angle}}{{Hypotenuse}}\]

In the given sum, we will get \[\sin \theta = \dfrac{{AB}}{{AC}}\]

Tangent: The ratio of side opposite to given angle and its adjacent side is called tangent. It is denoted as \[\tan \theta \].

\[\tan \theta = \dfrac{{Side\,opposite\,to\,given\,angle}}{{Side\,adjacent\,to\,given\,angle}}\]

In the given sum, we will get \[\tan \theta = \dfrac{{A'B'}}{{B'C'}}\]