Ch02. Inverse Trigonometric Functions

It's time for you to practice some questions.​

Instructions:

• The first four questions are exactly the same questions that I have discussed.
• Question 5 and 6 you should first attempt yourself. Solutions of Question 5 and 6 are given in the last video on this page, in case you need them.
• You might watch the practice video in case you need any help or instructions to continue with the practice.

Evaluate each of the following:

1. { \displaystyle sin^{-1} \left (-\sqrt{\frac{3}{2}} \right )}

2. { \displaystyle cot^{-1} \left ( - \sqrt{3} \right )}

3. { \displaystyle tan^{-1} \left ( {-\sqrt{3}} \right )}

4. { \displaystyle tan^{-1}(1)+cos^{-1} \left ( \frac{-1}{2} \right )+sin^{-1} \left ( \frac{-1}{2} \right )}

5. { \displaystyle cos^{-1} \left ( \frac{1}{2} \right )+2 \sin^{-1} \left ( \frac{1}{2} \right )}

6. { \displaystyle tan^{-1}\sqrt{3}-sec^{-1}(-2)}

Now, just watch the video to learn how to work with Question 7 and 8. We will practice them later.

Evaluate each of the following:
7. { \displaystyle sin^{-1} \left (\sin \left ( \frac{4 \pi}{5} \right ) \right )}
8. { \displaystyle sin^{-1} \left (\sin \left ( \frac{2 \pi}{3} \right ) \right )}

It's time for you to practice some questions.

Instructions:

• You might watch the practice video in case you need any help or instructions to continue with the practice.
• All the solutions of the following questions are given in the end videos of this page, in case you need them or to check your answers.

Evaluate each of the following:

7. { \displaystyle sin^{-1} \left (\sin \left ( \frac{4 \pi}{5} \right ) \right )}
8. { \displaystyle sin^{-1} \left (\sin \left ( \frac{2 \pi}{3} \right ) \right )}

9. { \displaystyle tan^{-1} \left (tan \left ( \frac{3 \pi}{4} \right ) \right )}
10. { \displaystyle cos^{-1} \left (cos \left ( \frac{7 \pi}{6} \right ) \right )}
11. { \displaystyle \sin \left ( \frac{\pi}{3}-sin^{-1} \left ( \frac{-1}{2} \right ) \right )}
12. { \displaystyle tan^{-1}\sqrt(3)-cot^{-1} \left (-\sqrt{3} \right )}
13. { \displaystyle csc^{-1}(-2)}

14. { \displaystyle sin^{-1} \left ( \sin \left [\frac{3 \pi}{5} \right ] \right ) }
16. { \displaystyle cos^{-1} \left ( cos \left ( \frac{13 \pi}{6} \right ) \right ) }
17. { \displaystyle tan^{-1} \left ( tan \left ( \frac{7 \pi}{6} \right ) \right ) }

18. { \displaystyle cos^{-1}[cos(-680°)] }
20. { \displaystyle tan^{-1} \left ( tan \left [ \frac{5 \pi}{6} \right ] \right ) }

Solutions

It's time for you to practice the questions that I have discussed.

Instructions:

• You might watch the practice video in case you need any help or instructions to continue with the practice.

Simplify:

1. { \displaystyle tan ^{-1} \left ( \sqrt{\frac{1-cos x}{1+cos x} } \right ) }

2. { \displaystyle tan ^{-1} \left (\frac{cos x-sin x}{cos x+sin x} \right )}

3. { \displaystyle cot ^{-1} \left (\frac{1}{\sqrt{x^{2} -1} } \right ), |x| >1}  

4. { \displaystyle tan ^{-1} \frac{1}{\sqrt{x^{2} -1} }, |x| >1}  

5. { \displaystyle tan ^{-1} \left (\frac{\sqrt{1+x^{2} } -1}{x} \right ), xne 0}  

6. { \displaystyle tan ^{-1} \frac{x}{\sqrt{a^{2} -x^{2} } }, |x|<a}  

7. { \displaystyle tan ^{-1} \left (\frac{3a^{2} x-x^{3} }{a^{3} -3ax^{2} } \right ), a>0; \frac{-a}{\sqrt{3} } \le x \le \frac{a}{\sqrt{3} } }  

First practice the following questions.

• The first seven questions are exactly the same questions that I have discussed. So, in case you need help you can watch the previous videos.
• Question 15, 16, 17 and 18 you should first attempt yourself. Their solutions are given above, in case you need them.
• You might watch the practice video in case you need any help or instructions to continue with the practice.

It's time for you to practice some questions.

Instructions:

• You might watch the practice video in case you need any help or instructions to continue with the practice.
• All the solutions of the following questions are given above, in case you need them or to check your answers.

Evaluate each of the following:

7. { \displaystyle sin^{-1} \left (\sin \left ( \frac{4 \pi}{5} \right ) \right )}
8. { \displaystyle sin^{-1} \left (\sin \left ( \frac{2 \pi}{3} \right ) \right )}

9. { \displaystyle tan^{-1} \left (tan \left ( \frac{3 \pi}{4} \right ) \right )}
10. { \displaystyle cos^{-1} \left (cos \left ( \frac{7 \pi}{6} \right ) \right )}
11. { \displaystyle \sin \left ( \frac{\pi}{3}-sin^{-1} \left ( \frac{-1}{2} \right ) \right )}
12. { \displaystyle tan^{-1}\sqrt(3)-cot^{-1} \left (-\sqrt{3} \right )}
13. { \displaystyle csc^{-1}(-2)}

14. { \displaystyle sin^{-1} \left ( \sin \left [\frac{3 \pi}{5} \right ] \right ) }
16. { \displaystyle cos^{-1} \left ( cos \left ( \frac{13 \pi}{6} \right ) \right ) }
17. { \displaystyle tan^{-1} \left ( tan \left ( \frac{7 \pi}{6} \right ) \right ) }

18. { \displaystyle cos^{-1}[cos(-680°)] }
20. { \displaystyle tan^{-1} \left ( tan \left [ \frac{5 \pi}{6} \right ] \right ) }

First practice the following questions.

Instructions:

• The first five questions are exactly the same questions that I have discussed. So, in case you need help you can watch the previous videos.
• Question 6, 7 and 8 you should first attempt yourself. Their solutions are given in the end, in case you need them.
• You might watch the practice video in case you need any help or instructions to continue with the practice.

Prove:

1. { \displaystyle 3sin ^{-1} x=sin ^{-1} (3x-4x^{3} ), x \in \left[-\frac{1}{2} ,\frac{1}{2} \right ]}
(NCERT Exercise 2.2 Q1)

2. { \displaystyle 3cos ^{-1} x=cos ^{-1} (4x^{3} -3x), x \in \left[\frac{1}{2} , 1\right ]}
(NCERT Exercise 2.2 Q2)

3. { \displaystyle tan ^{-1} \frac{2}{11} +tan ^{-1} \frac{7}{24} =tan ^{-1} \frac{1}{2} }
(NCERT Exercise 2.2 Q3)

4. { \displaystyle 2tan ^{-1} \frac{1}{2} +tan ^{-1} \frac{1}{7} =tan ^{-1} \frac{31}{17} }
(NCERT Exercise 2.2 Q4)

5. { \displaystyle tan ^{-1} \frac{3}{4} +tan ^{-1} \frac{3}{5} -tan ^{-1} \frac{8}{19} =\frac{pi }{4} }

6. { \displaystyle cot ^{-1} 7+cot ^{-1} 8+cot ^{-1} 18=cot ^{-1} 3}

7. { \displaystyle tan ^{-1} x+tan ^{-1} \left (\frac{2x}{1-x^{2} } \right )=tan ^{-1} \left (\frac{3x-x^{3} }{1-3x^{2} } \right )}
(NCERT Example 7)

8. { \displaystyle sin ^{-1} (2x\sqrt{1-x^{2} } )=2sin ^{-1} x=2cos ^{-1} x}
(NCERT Example 8, 9)

Solutions of Q6, Q7 & Q8

First practice the following questions.

Instructions:

• The first four questions are exactly the same questions that I have discussed. So, in case you need help you can watch the previous videos.
• Question 13, 14 and 15 you should first attempt yourself. Their solutions are given in the end, in case you need them.
• You might watch the practice video in case you need any help or instructions to continue with the practice.

Prove:

9. { \displaystyle tan ^{-1} \left (\frac{\sqrt{1+x^{2} } +\sqrt{1-x^{2} } }{\sqrt{1+x^{2} } -\sqrt{1-x^{2} } } \right )=\frac{{pi }}{2} -\frac{1}{2} sin ^{-1} x^{2} }

10. { \displaystyle tan ^{-1} \sqrt{x} =\frac{1}{2} cos ^{-1} \left (\frac{1-x}{1+x} \right ), x \in [01]}

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

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

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

Solutions of Q13, Q14 & Q15

First practice the following questions.

Instructions:

• The solutions are given in the end, in case you need them.
• You might watch the practice video in case you need any help or instructions to continue with the practice.

Prove:

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

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

18. { \displaystyle 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} }

Solutions of Q16, Q17 & Q18

First practice the following questions.

Instructions:

• The first three questions are exactly the same questions that I have discussed. So, in case you need help you can watch the previous videos.
• Question 22 and 23 you should first attempt yourself. Their solutions are given in the end, in case you need them.
• You might watch the practice video in case you need any help or instructions to continue with the practice.

Prove:

19. { \displaystyle sin ^{-1} \frac{3}{5} -sin ^{-1} \frac{8}{17} =cos ^{-1} \frac{84}{85} }

20. { \displaystyle sin ^{-1} \frac{8}{17} +sin ^{-1} \frac{3}{5} =tan ^{-1} \frac{77}{36} }

21. { \displaystyle tan ^{-1} \frac{63}{16} =sin ^{-1} \frac{5}{13} +cos ^{-1} \frac{3}{5} }

22. { \displaystyle cos ^{-1} \frac{4}{5} +cos ^{-1} \frac{12}{13} =cos ^{-1} \frac{33}{65} }

23. { \displaystyle cos ^{-1} \frac{12}{13} +sin ^{-1} \frac{3}{5} =sin ^{-1} \frac{56}{65} }

It's time for you to practice the questions that I have discussed.

Instructions:

• You might watch the practice video in case you need any help or instructions to continue with the practice.

Prove:

24. { \displaystyle sin ^{-1} \frac{12}{13} +cos ^{-1} \frac{4}{5} +tan ^{-1} \frac{63}{16} =pi }

25. { \displaystyle 2tan ^{-1} \left \{tan \frac{alpha }{2} tan \left (\frac{pi }{4} -\frac{beta }{2} \right )\right \}=tan ^{-1} \frac{sin \alpha cos \beta }{cos \alpha +sin \beta } }

26. { \displaystyle 2tan ^{-1} \left (\sqrt{\frac{a-b}{a+b} } tan \frac{\theta }{2} \right )=cos ^{-1} \left (\frac{acos \theta +b}{a+bcos \theta } \right )}

27. If { \displaystyle \sin \left(sin ^{-1} \frac{1}{5} +cos ^{-1} x \right )=1} , then find the value of x.

28. If { \displaystyle y=cot ^{-1} (\sqrt{cos x} )-tan ^{-1} (\sqrt{cos x} )} , prove that { \displaystyle sin y=tan ^{2} \frac{x}{2} }

First practice the following questions.

Instructions:

• The first eight questions are exactly the same questions that I have discussed. So, in case you need help you can watch the previous videos.
• Question 9, 10 and 11 you should first attempt yourself. Their solutions are given in the end, in case you need them.
• You might watch the practice video in case you need any help or instructions to continue with the practice.

Solve:

1. { \displaystyle tan ^{-1} \frac{x-1}{x-2} +tan ^{-1} \frac{x+1}{x+2} =\frac{pi }{4} }

2. { \displaystyle tan ^{-1} 2x+tan ^{-1} 3x=\frac{pi }{4} }

3. { \displaystyle tan ^{-1} (x+1)+tan ^{-1} (x-1)=tan ^{-1} \frac{8}{31} }

4. { \displaystyle 2tan ^{-1} (cos x)=tan ^{-1} (2, {csc}, x)}

5. { \displaystyle tan ^{-1} \frac{1-x}{1+x} =\frac{1}{2} tan ^{-1} x, (x>0)}

6. { \displaystyle sin ^{-1} (1-x)-2sin ^{-1} x=\frac{pi }{2} }

7. { \displaystyle sin ^{-1} x+sin ^{-1} (1-x)=cos ^{-1} x}

8. { \displaystyle sin ^{-1} 6x+sin ^{-1} 6\sqrt{3} x=-\frac{pi }{2} }

9. { \displaystyle tan ^{-1} (x-1)+tan ^{-1} x+tan ^{-1} (x+1)=tan ^{-1} 3x}

10. { \displaystyle 3sin ^{-1} \frac{2x}{1+x^{2} } -4cos ^{-1} \frac{1-x^{2} }{1+x^{2} } +2tan ^{-1} \frac{2x}{1-x^{2} } =\frac{pi }{3} }

11. If { \displaystyle sin ^{-1} \frac{2a}{1+a^{2} } -cos ^{-1} \frac{1-b^{2} }{1+b^{2} } =tan ^{-1} \frac{2x}{1-x^{2} } } , then prove that { \displaystyle x=\frac{a-b}{1+ab}.}

Solutions of Q9, Q10 & Q11

It's time for you to practice the questions that I have discussed.

Instructions:

• You might watch the practice video in case you need any help or instructions to continue with the practice.

Solve:

12. Evaluate: { \displaystyle tan ^{-1} \left (\frac{a+bx}{b-ax} \right ), x<\frac{b}{a} }

13. Prove: { \displaystyle tan ^{-1} \left (\frac{a-b}{1+ab} \right )+tan ^{-1} \left (\frac{b-c}{1+bc} \right )+tan ^{-1} \left (\frac{c-a}{1+ca} \right )=0}

14. If { \displaystyle tan ^{-1} x+tan ^{-1} y=\frac{4pi }{5} } , then find the value of { \displaystyle cot ^{-1} x+cot ^{-1}y} ?

15. If { \displaystyle tan ^{-1} \left (\frac{1}{1+1.2} \right )+tan ^{-1} \left (\frac{1}{1+2.3} \right )+...+tan ^{-1} \left (\frac{1}{1+n.(n+1)} \right )=tan ^{-1} phi } , then find the value of { \displaystyle phi} .

16. If { \displaystyle (tan ^{-1} x)^{2} +(cot ^{-1} x)^{2} =\frac{5pi ^{2} }{8} } , then find x.