Class 12: Inverse Trigonometric Functions – Exercise 4.10 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \text{Question 1:}$

$\displaystyle \text{(i) } \cot \Big( \sin^{-1} \frac{3}{4} + \sec^{-1} \frac{4}{3} \Big)$

$\displaystyle \text{(ii) } \sin \Big( \tan^{-1} x + \tan^{-1} \frac{1}{x} \Big) \text{ for } x < 0$

$\displaystyle \text{(iii) } \sin \Big( \tan^{-1} + \tan^{-1} \frac{1}{x} \Big) \text{ for } x > 0$

$\displaystyle \text{(iv) } \cot \Big( \tan^{-1} a + \cot^{-1} a \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2012] }$

$\displaystyle \text{(v) } \cos \Big( \sec^{-1} x + \mathrm{cosec}^{-1} x\Big), |x| \geq 1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) }$

$\displaystyle \cot \Big( \sin^{-1} \frac{3}{4} + \sec^{-1} \frac{4}{3} \Big)$

$\displaystyle = \cot \Big( \sin^{-1} \frac{3}{4} + \cos^{-1} \frac{3}{4} \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \sec^{-1} x = \cos^{-1} \frac{1}{x} \Big]$

$\displaystyle = \cot \Big( \frac{\pi}{2} \Big) = 0$

$\displaystyle \text{(ii) }$

$\displaystyle \sin \Big( \tan^{-1} x + \tan^{-1} \frac{1}{x} \Big)$

$\displaystyle = \sin \Big( \tan^{-1} (-x) + \tan^{-1} (-\frac{1}{x}) \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\because x < 0 ]$

$\displaystyle = \sin \Big( - \tan^{-1} (x) - \tan^{-1} (\frac{1}{x}) \Big)$

$\displaystyle = \sin \Big\{ - \Big[ \Big( \tan^{-1} (x) + \tan^{-1} (\frac{1}{x}) \Big) \Big] \Big\}$

$\displaystyle = \sin [ - ( \tan^{-1} x + \cot^{-1} x ) ] \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \tan^{-1} \frac{1}{x} = \cot^{-1} x \Big]$

$\displaystyle = -\sin ( \tan^{-1} x + \cot^{-1} x )$

$\displaystyle = - \sin \Big( \frac{\pi}{2} \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \Big]$

$\displaystyle = -1$

$\displaystyle \text{(iii) }$

$\displaystyle \sin \Big( \tan^{-1} + \tan^{-1} \frac{1}{x} \Big)$

$\displaystyle = \sin \Big( \tan^{-1} + \cot^{-1} x \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \tan^{-1} x = \cot^{-1} \frac{1}{x} \Big]$

$\displaystyle = \sin \Big( \frac{\pi}{2} \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \tan^{-1} x = \cot^{-1} x = \frac{\pi}{2} \Big]$

$\displaystyle = 1$

$\displaystyle \text{(iv) }$

$\displaystyle \cot \Big( \tan^{-1} a + \cot^{-1} a \Big)$

$\displaystyle = \cot \Big( \frac{\pi}{2} \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} \Big]$

$\displaystyle \text{(v) }$

$\displaystyle \cos \Big( \sec^{-1} x + \mathrm{cosec}^{-1} x\Big)$

$\displaystyle = \cos \Big( \frac{\pi}{2} \Big) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \sec^{-1} x + \mathrm{cosec}^{-1} x = \frac{\pi}{2} \Big]$

$\displaystyle = 0$

$\\$

$\displaystyle \text{Question 2. If} \cos^{-1} x + \cos^{-1} y = \frac{\pi}{4} , \text{ find the value of } \sin^{-1}x + \sin^{-1} y.$

$\displaystyle \text{Answer:}$

$\displaystyle \cos^{-1} x + \cos^{-1} y = \frac{\pi}{4}$

$\displaystyle \Rightarrow \frac{\pi}{2} - \sin^{-1} x + \frac{\pi}{2} - \sin^{-1} y = \frac{\pi}{4} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \Big]$

$\displaystyle \Rightarrow \pi - ( \sin^{-1} x + \sin^{-1} y)= \frac{\pi}{4}$

$\displaystyle \Rightarrow \sin^{-1} x + \sin^{-1} y = \frac{3\pi}{4}$

$\\$

$\displaystyle \text{Question 3. If} \sin^{-1} x + \sin^{-1} y = \frac{\pi}{3} , \text{ and } \cos^{-1} x - \cos^{-1} y = \frac{\pi}{6} , \text{ find the value of } \\ x \text{ and } y$

$\displaystyle \text{Answer:}$

$\displaystyle \cos^{-1} x - \cos^{-1} y = \frac{\pi}{6}$

$\displaystyle \Rightarrow \frac{\pi}{2} - \sin^{-1} x - \frac{\pi}{2} + \sin^{-1} y = \frac{\pi}{6} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \Big]$

$\displaystyle \Rightarrow -(\sin^{-1} x - \sin^{-1} y ) = \frac{\pi}{6}$

$\displaystyle \Rightarrow \sin^{-1} x - \sin^{-1} y = -\frac{\pi}{6}$

$\displaystyle \text{Solving } \\ \\ \sin^{-1} x + \sin^{-1} y = \frac{\pi}{3} \text{ and } \sin^{-1} x - \sin^{-1} y = -\frac{\pi}{6} \text{ we will get } 2 \sin^{-1} x = \frac{\pi}{6}$

$\displaystyle \Rightarrow \sin^{-1} x = \frac{\pi}{12}$

$\displaystyle \Rightarrow x = \sin \frac{\pi}{12} = \frac{\sqrt{3} -1 }{2 \sqrt{2}}$

$\displaystyle \text{and}$

$\displaystyle \sin^{-1} y = \frac{\pi}{3} - \sin^{-1} x$

$\displaystyle \Rightarrow \sin^{-1} y = \frac{\pi}{3} - \frac{\pi}{12}$

$\displaystyle \Rightarrow \sin^{-1} y = \frac{\pi}{4}$

$\displaystyle \Rightarrow y = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$

$\\$

$\displaystyle \text{Question 4. If} \cot \Big( \cos^{-1} \frac{3}{5} + \sin^{-1} x \Big) = 0 , \text{ find the value of } x$

$\displaystyle \text{Answer:}$

$\displaystyle \cot \Big( \cos^{-1} \frac{3}{5} + \sin^{-1} x \Big) = 0$

$\displaystyle \Rightarrow \cos^{-1} \frac{3}{5} + \sin^{-1} x = \cot 0$

$\displaystyle \Rightarrow \cos^{-1} \frac{3}{5} + \sin^{-1} x = \frac{\pi}{2}$

$\displaystyle \Rightarrow \cos^{-1} \frac{3}{5} = \frac{\pi}{2} - \sin^{-1} x$

$\displaystyle \Rightarrow \cos^{-1} \frac{3}{5} =\cos^{-1} x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \Big]$

$\displaystyle \Rightarrow x = \frac{3}{5}$

$\\$

$\displaystyle \text{Question 5. If } (\sin^{-1} x)^2 + (\cos^{-1} x)^2 = \frac{17\pi^2}{36}, \text{ find } x.$

$\displaystyle \text{Answer:}$

$\displaystyle (\sin^{-1} x)^2 + (\cos^{-1} x)^2 = \frac{17\pi^2}{36}$

$\displaystyle (\sin^{-1} x)^2 + \Big(\frac{\pi}{2} - \sin^{-1} x \Big)^2 = \frac{17\pi^2}{36}$

$\displaystyle \text{Let } \sin^{-1} x = y$

$\displaystyle \therefore y^2 + \Big(\frac{\pi}{2} - y \Big)^2 = \frac{17\pi^2}{36}$

$\displaystyle \Rightarrow y^2 + \frac{\pi^2}{4} + y^2 - 2 \times \frac{\pi}{2} \times y = \frac{17\pi^2}{36}$

$\displaystyle \Rightarrow 2y^2 - \pi y = \frac{2 \pi^2}{9}$

$\displaystyle \Rightarrow 18y^2 = 9 \pi y - 2 \pi^2 = 0$

$\displaystyle \Rightarrow 6y(3y - 2 \pi) + \pi ( 3y - 2 \pi) = 0$

$\displaystyle \Rightarrow ( 3y - 2 \pi) ( 6y + \pi) = 0$

$\displaystyle \Rightarrow y = - \frac{\pi}{6} \ \ \ \ \ \ \ \ \ \ \text{[Neglecting } y = \frac{2}{3} \pi \text{ as it is not satisfying the equation] }$

$\displaystyle \therefore x = \sin y = \sin \Big(-\frac{\pi}{6} \Big) = - \frac{1}{2}$

$\\$

$\displaystyle \text{Question 6. Solve: } \sin \Big\{ \sin^{-1} \frac{1}{5} + \cos^{-1} x\Big\} = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[CBSE 2014] }$

$\displaystyle \text{Answer:}$

$\displaystyle \sin \Big\{ \sin^{-1} \frac{1}{5} + \cos^{-1} x\Big\} = 1$

$\displaystyle \Rightarrow \Big\{ \sin^{-1} \frac{1}{5} + \cos^{-1} x\Big\} = \sin^{-1} 1$

$\displaystyle \Rightarrow \sin^{-1} \frac{1}{5} + \cos^{-1} x = \frac{\pi}{2}$

$\displaystyle \Rightarrow \sin^{-1} \frac{1}{5} = \frac{\pi}{2} - \cos^{-1} x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \sin^{-1} x = \frac{\pi}{2} - \cos^{-1} x \Big]$

$\displaystyle \Rightarrow \sin^{-1} \frac{1}{5} = \sin^{-1} x$

$\displaystyle \Rightarrow x = \frac{1}{5}$

$\\$

$\displaystyle \text{Question 7. Solve: } \sin^{-1} x = \frac{\pi}{6} + \cos^{-1} x$

$\displaystyle \text{Answer:}$

$\displaystyle \sin^{-1} x= \frac{\pi}{6} + \cos^{-1} x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \Big]$

$\displaystyle \Rightarrow \sin^{-1} x= \frac{\pi}{6} + \frac{\pi}{2} - \sin^{-1} x$

$\displaystyle \Rightarrow 2 \sin^{-1}x = \frac{2\pi}{3}$

$\displaystyle \Rightarrow \sin^{-1} x= \frac{\pi}{3}$

$\displaystyle \Rightarrow x = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$

$\\$

$\displaystyle \text{Question 8. Solve: } 4 \sin^{-1} x = \pi - \cos^{-1} x$

$\displaystyle \text{Answer:}$

$\displaystyle 4 \sin^{-1} x = \pi - \cos^{-1} x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \Big]$

$\displaystyle \Rightarrow 4 \sin^{-1} x = \pi - \Big( \frac{\pi}{2} - \sin^{-1} x \Big)$

$\displaystyle \Rightarrow 4 \sin^{-1} x = \frac{\pi}{2} + \sin^{-1} x$

$\displaystyle \Rightarrow 3 \sin^{-1} x = \frac{\pi}{2}$

$\displaystyle \Rightarrow \sin^{-1} x = \frac{\pi}{3}$

$\displaystyle \Rightarrow x = \sin \frac{\pi}{6} = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 9. Solve: } \tan^{-1} x + 2 \cot^{-1} x = \frac{2\pi}{3}$

$\displaystyle \text{Answer:}$

$\displaystyle \tan^{-1} x + 2 \cot^{-1} x = \frac{2\pi}{3} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \Big]$

$\displaystyle \Rightarrow \tan^{-1} x + 2 \Big( \frac{\pi}{2} - \tan^{-1} x \Big) = \frac{2\pi}{3}$

$\displaystyle \Rightarrow \tan^{-1} x + \pi - 2 \tan^{-1} x = \frac{2\pi}{3}$

$\displaystyle \Rightarrow \tan^{-1} x = \frac{\pi}{3}$

$\displaystyle \Rightarrow x = \tan \frac{\pi}{3} = \sqrt{3}$

$\\$

$\displaystyle \text{Question 10. Solve: } 5 \tan^{-1} x + 3 \cot^{-1} x = 2\pi$

$\displaystyle \text{Answer:}$

$\displaystyle 5 \tan^{-1} x + 3 \cot^{-1} x = 2\pi \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Big[ \because \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x \Big]$

$\displaystyle \Rightarrow 5 \tan^{-1} x + 3 \Big( \frac{\pi}{2} - \tan^{-1} x \Big) = 2\pi$

$\displaystyle \Rightarrow 5 \tan^{-1} x + 3\frac{\pi}{2} - 3 \tan^{-1} x = 2\pi$

$\displaystyle \Rightarrow 2 \tan^{-1} x = \frac{\pi}{2}$

$\displaystyle \Rightarrow \tan^{-1} x = \frac{\pi}{4}$

$\displaystyle \Rightarrow x = \tan \frac{\pi}{4} = 1$

$\\$

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Class 12: Inverse Trigonometric Functions – Exercise 4.10

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