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  • Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 3 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Inverse Trigonometric Function Exercise 3.10 Question 3 Maths Textbook Solution.

Answers (1)

Answer:x= \frac{\sqrt{3-1}}{2\sqrt{2}}\: and\: y= \frac{1}{\sqrt{2}}
Hint: Changing \sin^{-1}x to\cos^{-1}x and vice versa; We know that,
\sin^{-1}x+\cos^{-1}x= \frac{\pi }{2}
Given:\cos^{-1}x-\cos^{-1}y  = \frac{\pi }{6}  and\sin^{-1}x+ \sin^{-1}y = \frac{\pi }{3}
Here, we have to compute x  and y.
Solution:
Let’s replace \cos^{-1}x-\cos^{-1}y  by \sin^{-1}x\: and\: \sin^{-1}y , therefore
\! \! \! \! \! \! \! \! \Rightarrow \cos^{-1}x-\cos^{-1}y = \frac{\pi }{6} \cdot \cdot \cdot (1)\\ \Rightarrow \sin^{-1}x+ \sin^{-1}y = \frac{\pi }{3} \cdot \cdot \cdot \cdot (2)\\ \Rightarrow \frac{\pi }{2} - \sin^{-1}x - \frac{\pi }{2} + \sin^{-1}y = \frac{\pi }{6}\\ -(\sin^{-1}x-\sin^{-1}y) = \frac{\pi }{6} \cdot \cdot \cdot (3)
Subtracting Equation 3 from 2,
\! \! \! \! \! \! \! \! \! \sin^{-1}x+\sin^{-1}y +\sin^{-1}x-\sin^{-1}y = \frac{\pi }{3} - \frac{\pi }{6}\\ 2 \sin^{-1}x = \frac{\pi }{6}\\ \sin^{-1}x = \frac{\pi }{12}\\ x = \sin \frac{\pi }{12}\\
x= \frac{\sqrt{3}-1}{2\sqrt{2}}
Adding Equation 2 and 3,
\! \! \! \! \! \! \! \! \! 2 \sin^{-1}y = \frac{\pi }{3} + \frac{\pi }{6}\\ 2 \sin^{-1}y = \frac{\pi }{2}\\ \sin^{-1}y = \frac{\pi }{4}\\ y = \sin \frac{\pi }{4}\\ y = \frac{1}{\sqrt2}
Concept: Properties and relations between inverse trigonometric functions.
Note: Values of Trigonometric functions for various radians and degrees.    

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