
Find the correct option.
The value of \[\sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)] + \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\] is
A. 1
B. 2
C. 3
D. 4
Answer
232.8k+ views
Hint: The subparts of the given expression will be first defined and using relevant inverse trigonometric identities, those subparts will be simplified and then putting the sine and cosine values of the simplified angles, the final value of the expression will be calculated and the correct option will be chosen.
Formulae Used: The following inverse trigonometric identities will be used to simplify the given expression:
1. \[{\sec ^{ - 1}}(x) = {\cos ^{ - 1}}(\dfrac{1}{x})\]
2. \[{\tan ^{ - 1}}(x) = {\cot ^{ - 1}}(\dfrac{1}{x})\]
3. \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]
4. \[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}\]
Complete step-by-step solution:
We have been given the expression \[\sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)] + \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\].
We have to simplify the above expression and find its value to choose the correct option.
Let, \[y = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)] + \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\]
We will define the sub-parts of the expression as follows:
Let, \[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)]\]
And \[{y_2} = \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\]
So, \[y = {y_1} + {y_2}\]
First, we will simplify \[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)]\] as follows:
We know that \[{\sec ^{ - 1}}(x) = {\cos ^{ - 1}}(\dfrac{1}{x})\]
Substituting \[x\] with \[3\], we have
\[{\sec ^{ - 1}}(3) = {\cos ^{ - 1}}(\dfrac{1}{3})\]
So, \[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)]\]
\[ = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\cos ^{ - 1}}(\dfrac{1}{3})]\] [Since \[{\sec ^{ - 1}}(3) = {\cos ^{ - 1}}(\dfrac{1}{3})\]]
Now, using the formula \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\] and substituting \[x\] with \[\dfrac{1}{3}\], we have
\[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\cos ^{ - 1}}(\dfrac{1}{3})]\]
\[ = \sin \dfrac{\pi }{2}\]
Similarly, we will simplify \[{y_2} = \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\] as follows:
We know \[{\tan ^{ - 1}}(x) = {\cot ^{ - 1}}(\dfrac{1}{x})\]
Substituting \[x\] with \[\dfrac{1}{2}\], we have
\[{\tan ^{ - 1}}(\dfrac{1}{2}) = {\cot ^{ - 1}}(2)\] [Since \[\dfrac{1}{{\dfrac{1}{2}}} = 1 \times \dfrac{2}{1} = 2\]]
So, \[{y_2} = \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\]
\[ = \cos [{\cot ^{ - 1}}(2) + {\tan ^{ - 1}}(2)]\] [Since \[{\tan ^{ - 1}}(\dfrac{1}{2}) = {\cot ^{ - 1}}(2)\] ]
\[ = \cos [{\tan ^{ - 1}}(2) + {\cot ^{ - 1}}(2)]\]
Now, using the formula \[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}\] and substituting \[x\] with \[2\], we have
\[{y_2} = \cos [{\tan ^{ - 1}}(2) + {\cot ^{ - 1}}(2)]\]
\[ = \cos \dfrac{\pi }{2}\]
Now, assigning the sine and cosine values of the angle \[\dfrac{\pi }{2}\] , we will find the value of \[y\] as follows:
\[y = {y_1} + {y_2}\]
\[ = \sin \dfrac{\pi }{2} + \cos \dfrac{\pi }{2}\] [Since \[\sin \dfrac{\pi }{2} = 1\] and \[\cos \dfrac{\pi }{2} = 0\] ]
\[ = 1 + 0\]
\[ = 1\]
So, the value of \[\sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)] + \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\] is equal to \[1\] .
Hence, option A. 1 is the correct answer.
Note: The inverse trigonometric functions are the inverse functions of the trigonometric functions. Especially, those are the inverse functions of the sine, cosine, tangent, cotangent, secant and cosecant functions and are used to obtain an angle from any of the angle’s trigonometric ratios. The inverse symbol, for example, \[{\sin ^{ - 1}}x\] should not be confused with \[{(\sin x)^{ - 1}}\] .
Formulae Used: The following inverse trigonometric identities will be used to simplify the given expression:
1. \[{\sec ^{ - 1}}(x) = {\cos ^{ - 1}}(\dfrac{1}{x})\]
2. \[{\tan ^{ - 1}}(x) = {\cot ^{ - 1}}(\dfrac{1}{x})\]
3. \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\]
4. \[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}\]
Complete step-by-step solution:
We have been given the expression \[\sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)] + \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\].
We have to simplify the above expression and find its value to choose the correct option.
Let, \[y = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)] + \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\]
We will define the sub-parts of the expression as follows:
Let, \[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)]\]
And \[{y_2} = \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\]
So, \[y = {y_1} + {y_2}\]
First, we will simplify \[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)]\] as follows:
We know that \[{\sec ^{ - 1}}(x) = {\cos ^{ - 1}}(\dfrac{1}{x})\]
Substituting \[x\] with \[3\], we have
\[{\sec ^{ - 1}}(3) = {\cos ^{ - 1}}(\dfrac{1}{3})\]
So, \[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)]\]
\[ = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\cos ^{ - 1}}(\dfrac{1}{3})]\] [Since \[{\sec ^{ - 1}}(3) = {\cos ^{ - 1}}(\dfrac{1}{3})\]]
Now, using the formula \[{\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}\] and substituting \[x\] with \[\dfrac{1}{3}\], we have
\[{y_1} = \sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\cos ^{ - 1}}(\dfrac{1}{3})]\]
\[ = \sin \dfrac{\pi }{2}\]
Similarly, we will simplify \[{y_2} = \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\] as follows:
We know \[{\tan ^{ - 1}}(x) = {\cot ^{ - 1}}(\dfrac{1}{x})\]
Substituting \[x\] with \[\dfrac{1}{2}\], we have
\[{\tan ^{ - 1}}(\dfrac{1}{2}) = {\cot ^{ - 1}}(2)\] [Since \[\dfrac{1}{{\dfrac{1}{2}}} = 1 \times \dfrac{2}{1} = 2\]]
So, \[{y_2} = \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\]
\[ = \cos [{\cot ^{ - 1}}(2) + {\tan ^{ - 1}}(2)]\] [Since \[{\tan ^{ - 1}}(\dfrac{1}{2}) = {\cot ^{ - 1}}(2)\] ]
\[ = \cos [{\tan ^{ - 1}}(2) + {\cot ^{ - 1}}(2)]\]
Now, using the formula \[{\tan ^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}\] and substituting \[x\] with \[2\], we have
\[{y_2} = \cos [{\tan ^{ - 1}}(2) + {\cot ^{ - 1}}(2)]\]
\[ = \cos \dfrac{\pi }{2}\]
Now, assigning the sine and cosine values of the angle \[\dfrac{\pi }{2}\] , we will find the value of \[y\] as follows:
\[y = {y_1} + {y_2}\]
\[ = \sin \dfrac{\pi }{2} + \cos \dfrac{\pi }{2}\] [Since \[\sin \dfrac{\pi }{2} = 1\] and \[\cos \dfrac{\pi }{2} = 0\] ]
\[ = 1 + 0\]
\[ = 1\]
So, the value of \[\sin [{\sin ^{ - 1}}(\dfrac{1}{3}) + {\sec ^{ - 1}}(3)] + \cos [{\tan ^{ - 1}}(\dfrac{1}{2}) + {\tan ^{ - 1}}(2)]\] is equal to \[1\] .
Hence, option A. 1 is the correct answer.
Note: The inverse trigonometric functions are the inverse functions of the trigonometric functions. Especially, those are the inverse functions of the sine, cosine, tangent, cotangent, secant and cosecant functions and are used to obtain an angle from any of the angle’s trigonometric ratios. The inverse symbol, for example, \[{\sin ^{ - 1}}x\] should not be confused with \[{(\sin x)^{ - 1}}\] .
Recently Updated Pages
The area of an expanding rectangle is increasing at class 12 maths JEE_Main

If y xxx cdots infty then find dfracdydx A yxy 1 B class 12 maths JEE_Main

Area vs Volume: Key Differences Explained for Students

Mutually Exclusive vs Independent Events: Key Differences Explained

Geometry of Complex Numbers Explained

JEE General Topics in Chemistry Important Concepts and Tips

Trending doubts
JEE Main 2026: Admit Card Out, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Understanding Electromagnetic Waves and Their Importance

Inductive Effect and Its Role in Acidic Strength

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

