
Solve $2{\sin ^{ - 1}}(\dfrac{3}{5}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}})$
A. $\dfrac{\pi }{2}$
B. $\dfrac{{2\pi }}{3}$
C. $\dfrac{{5\pi }}{3}$
D. None of these
Answer
219.9k+ views
Hint: It is necessary to remember a few fundamental formulas to solve trigonometric equations. To answer this particular question, we need to remember the formulas of inverse trigonometric ratios in addition to the usual trigonometric formulas.
Formula Used:
${\sin ^2}x + {\cos ^2}x = 1$
$\cos 2A = 2{\cos ^2}A - 1$
Complete step by step Solution:
Let us suppose that $2{\sin ^{ - 1}}(\dfrac{3}{5}) = A$
Then we have, ${\sin ^{ - 1}}(\dfrac{3}{5}) = \dfrac{A}{2}$
$ \Rightarrow \sin \dfrac{A}{2} = \dfrac{3}{5}$
Since ${\sin ^2}x + {\cos ^2}x = 1$
$ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$
$ \Rightarrow {\cos ^2}\left( {\dfrac{A}{2}} \right) = 1 - {\sin ^2}\left( {\dfrac{A}{2}} \right)$
$ = 1 - {\left( {\dfrac{3}{5}} \right)^2}$ $ = 1 - \dfrac{9}{{25}}$
$ \Rightarrow \cos \dfrac{A}{2} = \sqrt {(1 - \dfrac{9}{{25}})} $
$ \Rightarrow \cos \dfrac{A}{2} = \sqrt {\dfrac{{16}}{{25}}} = \dfrac{4}{5}$
Now we will find the value of $\cos A$
We know that, $\cos 2A = 2{\cos ^2}A - 1$
$ \Rightarrow \cos A = 2{\cos ^2}(\dfrac{A}{2}) - 1$
Therefore, we have, $\cos A = 2(\dfrac{{16}}{{25}}) - 1$
$ \Rightarrow \cos A = \dfrac{7}{{25}}$
Thus, $A = {\cos ^{ - 1}}(\dfrac{7}{{25}})$
Now,
$2{\sin ^{ - 1}}(\dfrac{3}{5}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}}) = {\cos ^{ - 1}}(\dfrac{7}{{25}}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}})$
Using the formula ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}[xy - \sqrt {(1 - {x^2})} \sqrt {(1 - {y^2})} ]$
${\cos ^{ - 1}}(\dfrac{7}{{25}}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}}) = {\cos ^{ - 1}}[(\dfrac{7}{{25}})(\dfrac{{24}}{{25}}) - \sqrt {1 - \dfrac{{49}}{{625}}} \sqrt {1 - \dfrac{{576}}{{625}}} ]$
$ = {\cos ^{ - 1}}[\dfrac{{168}}{{625}} - \sqrt {\dfrac{{576}}{{625}}} \sqrt {\dfrac{{49}}{{625}}} ]$
On further solving we get,
${\cos ^{ - 1}}(\dfrac{7}{{25}}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}}) = {\cos ^{ - 1}}[\dfrac{{168}}{{625}} - (\dfrac{{24}}{{25}})(\dfrac{7}{{25}})]$
$ = {\cos ^{ - 1}}[\dfrac{{168}}{{625}} - \dfrac{{168}}{{625}}]$
$ = {\cos ^{ - 1}}0$
$ = {\cos ^{ - 1}}(\cos \dfrac{\pi }{2})$
$ = \dfrac{\pi }{2}$
Hence, the correct option is (A).
Note: Assume the value ${\sin ^{ - 1}}\dfrac{3}{5}$ with some suitable constant for easy calculation. After solving the question, in the end, we need to replace the variable we have taken with the inverse of sin which will give us the final answer.
Formula Used:
${\sin ^2}x + {\cos ^2}x = 1$
$\cos 2A = 2{\cos ^2}A - 1$
Complete step by step Solution:
Let us suppose that $2{\sin ^{ - 1}}(\dfrac{3}{5}) = A$
Then we have, ${\sin ^{ - 1}}(\dfrac{3}{5}) = \dfrac{A}{2}$
$ \Rightarrow \sin \dfrac{A}{2} = \dfrac{3}{5}$
Since ${\sin ^2}x + {\cos ^2}x = 1$
$ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$
$ \Rightarrow {\cos ^2}\left( {\dfrac{A}{2}} \right) = 1 - {\sin ^2}\left( {\dfrac{A}{2}} \right)$
$ = 1 - {\left( {\dfrac{3}{5}} \right)^2}$ $ = 1 - \dfrac{9}{{25}}$
$ \Rightarrow \cos \dfrac{A}{2} = \sqrt {(1 - \dfrac{9}{{25}})} $
$ \Rightarrow \cos \dfrac{A}{2} = \sqrt {\dfrac{{16}}{{25}}} = \dfrac{4}{5}$
Now we will find the value of $\cos A$
We know that, $\cos 2A = 2{\cos ^2}A - 1$
$ \Rightarrow \cos A = 2{\cos ^2}(\dfrac{A}{2}) - 1$
Therefore, we have, $\cos A = 2(\dfrac{{16}}{{25}}) - 1$
$ \Rightarrow \cos A = \dfrac{7}{{25}}$
Thus, $A = {\cos ^{ - 1}}(\dfrac{7}{{25}})$
Now,
$2{\sin ^{ - 1}}(\dfrac{3}{5}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}}) = {\cos ^{ - 1}}(\dfrac{7}{{25}}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}})$
Using the formula ${\cos ^{ - 1}}x + {\cos ^{ - 1}}y = {\cos ^{ - 1}}[xy - \sqrt {(1 - {x^2})} \sqrt {(1 - {y^2})} ]$
${\cos ^{ - 1}}(\dfrac{7}{{25}}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}}) = {\cos ^{ - 1}}[(\dfrac{7}{{25}})(\dfrac{{24}}{{25}}) - \sqrt {1 - \dfrac{{49}}{{625}}} \sqrt {1 - \dfrac{{576}}{{625}}} ]$
$ = {\cos ^{ - 1}}[\dfrac{{168}}{{625}} - \sqrt {\dfrac{{576}}{{625}}} \sqrt {\dfrac{{49}}{{625}}} ]$
On further solving we get,
${\cos ^{ - 1}}(\dfrac{7}{{25}}) + {\cos ^{ - 1}}(\dfrac{{24}}{{25}}) = {\cos ^{ - 1}}[\dfrac{{168}}{{625}} - (\dfrac{{24}}{{25}})(\dfrac{7}{{25}})]$
$ = {\cos ^{ - 1}}[\dfrac{{168}}{{625}} - \dfrac{{168}}{{625}}]$
$ = {\cos ^{ - 1}}0$
$ = {\cos ^{ - 1}}(\cos \dfrac{\pi }{2})$
$ = \dfrac{\pi }{2}$
Hence, the correct option is (A).
Note: Assume the value ${\sin ^{ - 1}}\dfrac{3}{5}$ with some suitable constant for easy calculation. After solving the question, in the end, we need to replace the variable we have taken with the inverse of sin which will give us the final answer.
Recently Updated Pages
Geometry of Complex Numbers Explained

Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Trending doubts
Understanding Atomic Structure for Beginners

Understanding Centrifugal Force in Physics

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

Understanding Electromagnetic Waves and Their Importance

Understanding Entropy Changes in Different Processes

Common Ion Effect: Concept, Applications, and Problem-Solving

Other Pages
Understanding Excess Pressure Inside a Liquid Drop

Understanding Geostationary and Geosynchronous Satellites

Understanding Elastic Collisions in Two Dimensions

Devuthani Ekadashi 2025: Correct Date, Shubh Muhurat, Parana Time & Puja Vidhi

Quadratic Equation Questions with Solutions & PDF Practice Sets

Difference Between Exothermic and Endothermic Reactions: Key Differences, Examples & Diagrams

