Question

The value of \[{{\cot }^{-1}}\left( 9 \right)\]+ \[\cos e{{c}^{-1}}\dfrac{\sqrt{41}}{4}\] is
(a) \[\dfrac{\pi }{2}\]
(b) \[\dfrac{\pi }{4}\]
(c ) \[\dfrac{\pi }{3}\]
(d) \[\pi \]

Answer 1

Hint: This question is based on inverse trigonometric functions. To solve this question, first, we have the proper knowledge about all inverse trigonometric functions, then we are able to solve the question. Inverse trigonometric functions are based on basic Trigonometric functions. First, we find the equation which is related to our given question, then by solving and with the help of more trigonometric functions which satisfy our equation, we get our desirable answer.

Formula Used:
${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$
${{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}$
${{\sec }^{-1}}x+\cos e{{c}^{-1}}x=\dfrac{\pi }{2}$

Complete step by step Solution:
We use them according to the given equation.
We have to find the value of \[{{\cot }^{-1}}\left( 9 \right)\] + \[\cos e{{c}^{-1}}\dfrac{\sqrt{41}}{4}\]. . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . (1)
We know $\cos e{{c}^{-1}}x={{\cot }^{-1}}\sqrt{{{x}^{2}}-1}$
\[\cos e{{c}^{-1}}\dfrac{\sqrt{41}}{4}\] = ${{\cot }^{-1}}\sqrt{\dfrac{41}{16}-1}$
By solving the above equation, we get
${{\cot }^{-1}}\sqrt{\dfrac{41}{16}-1}$= ${{\cot }^{-1}}\dfrac{5}{4}$
By putting this value in equation (1), we get
\[{{\cot }^{-1}}\left( 9 \right)\] + ${{\cot }^{-1}}\dfrac{5}{4}$
We know ${{\cot }^{-1}}x=\dfrac{1}{{{\tan }^{-1}}x}$
Then we get ${{\tan }^{-1}}\left( \dfrac{1}{9} \right)+{{\tan }^{-1}}\left( \dfrac{4}{5} \right)$
We know the formula of ${{\tan }^{-1}}(x)+{{\tan }^{-1}}(y)={{\tan }^{-1}}\left( \dfrac{x+y}{1+xy} \right)$
Where x = $\dfrac{1}{9}$and y = $\dfrac{4}{5}$
Put the value of ${{\tan }^{-1}}\left( \dfrac{1}{9} \right)+{{\tan }^{-1}}\left( \dfrac{4}{5} \right)$ in the above formula, we get
${{\tan }^{-1}}\left( \dfrac{1}{9} \right)+{{\tan }^{-1}}\left( \dfrac{4}{5} \right)$ = ${{\tan }^{-1}}\left( \dfrac{\dfrac{1}{9}+\dfrac{4}{5}}{1-\dfrac{1}{9}\times \dfrac{4}{5}} \right)$
By taking the LCM and solving the above equation, we get
${{\tan }^{-1}}\left( \dfrac{\dfrac{5+9}{45}}{1-\dfrac{4}{45}} \right)$ = ${{\tan }^{-1}}\left( 1 \right)$
We know value of $\tan \left( \dfrac{\pi }{4} \right)$ = 1
${{\tan }^{-1}}\left( 1 \right)$= ${{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)$
That means ${{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)$ = $\dfrac{\pi }{4}$
Value of \[{{\cot }^{-1}}\left( 9 \right)\] + \[\cos e{{c}^{-1}}\dfrac{\sqrt{41}}{4}\] = $\dfrac{\pi }{4}$

Hence, the correct option is b.

Note: To solve these types of questions, it is very important to know the inverse trigonometric equations. Also, it is important to know how to put the values into the equations. If we don’t know the trigonometric equations, we are not able to solve the question. These equations are based on basic trigonometric formulas. If we have the proper knowledge and deep understanding of trigonometric functions, it is not difficult to learn inverse trigonometric equations.