Practice Questions Inverse Trigonometry Lecture 4 Part 2

Part - 2 Lecture - 4 Chapter 2 Inverse Trigonometric Functions

“The only place where success comes before work is in the dictionary.” –Vidal Sassoon

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Questions discussed in this lecture

Prove:
11. \frac{9\pi }{8} -\frac{9}{4} \sin ^{-1} \frac{1}{3} =\frac{9}{4} \sin ^{-1} \frac{2\sqrt{2} }{3}
(NCERT Miscellaneous Exercise Q12)

12. \tan \left(\frac{\pi }{4} +\frac{1}{2} \cos ^{-1} \frac{a}{b} \right)+\tan \left(\frac{\pi }{4} -\frac{1}{2} \cos ^{-1} \frac{a}{b} \right)=\frac{2b}{a}

13. \sin [\cot ^{-1} { \cos (\tan ^{-1} x)} ]=\frac{\sqrt{x^{2} +1} }{\sqrt{x^{2} +2} }

14. \tan ^{-1} \left(\frac{x}{\sqrt{a^{2} -x^{2} } } \right)=\sin ^{-1} \frac{x}{a} =\cot ^{-1} \left(\frac{\sqrt{a^{2} -x^{2} } }{a} \right)

15. \tan ^{-1} \left(\frac{m}{n} \right)-\tan ^{-1} \left(\frac{m-n}{m+n} \right)=\frac{\pi }{4}

16. \tan ^{-1} \frac{1}{5} +\tan ^{-1} \frac{1}{7} +\tan ^{-1} \frac{1}{3} +\tan ^{-1} \frac{1}{8} =\frac{\pi }{4}

(NCERT Miscellaneous Exercise Q8)

17. 4\tan ^{-1} \frac{1}{5} -\tan ^{-1} \frac{1}{70} +\tan ^{-1} \frac{1}{99} =\frac{\pi }{4}

18. \tan ^{-1} \left(\frac{\cos x}{1-\sin x} \right)-\cot ^{-1} \left(\sqrt{\frac{1+\cos x}{1-\cos x} } \right)=\frac{\pi }{4}

19. \sin ^{-1} \frac{3}{5} -\sin ^{-1} \frac{8}{17} =\cos ^{-1} \frac{84}{85}
(NCERT Example 10)

20. \sin ^{-1} \frac{8}{17} +\sin ^{-1} \frac{3}{5} =\tan ^{-1} \frac{77}{36}
(NCERT Miscellaneous Exercise Q4)

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