Inverse Trigonometric Functions Part - 2

Part 2 of this course builds upon the basics learned in Part 1 and takes your understanding of inverse trigonometric functions to the next level. This course is recorded in Hindi for English medium syllabus of Class 12 Mathematics.
Part 2 of this course covers advanced topics in inverse trigonometric functions such as proving identities and solving complex problems using Higher Order Thinking Skills. You will learn how to apply the concepts of inverse trigonometric functions to real-world scenarios and solve problems using techniques such as substitution, factorization, and squaring. With detailed solutions and explanations for each question, this course is perfect for students who want to excel in their Class 12 Mathematics exams or competitive exams like JEE Main, JEE Advanced, NEET, etc. By the end of this course, you will have a thorough understanding of inverse trigonometric functions and be able to solve even the most challenging problems with ease.

4 hours 26 minutes Course Duration
English Syllabus medium
Hindi + English Explanation
Not Logged in Enrollment status
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Course Content

Lecture – 1

00:41:19

Questions Discussed:

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

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

Question 3. { \displaystyle {{\tan }^{-1}}\frac{2}{11}} + { \displaystyle {{\tan }^{-1}}\frac{7}{24} } = { \displaystyle{{\tan }^{-1}}\frac{1}{2} }

Question 4. { \displaystyle 2{{\tan }^{-1}}\frac{1}{2}} + { \displaystyle{{\tan }^{-1}}\frac{1}{7}} = { \displaystyle{{\tan }^{-1}}\frac{31}{17} }

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

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

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

Question 8. { \displaystyle {{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}}) } = { \displaystyle 2{{\sin }^{-1}}x} = { \displaystyle 2{{\cos }^{-1}}x }

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

Question 10 { \displaystyle {{\tan }^{-1}}\sqrt{x} } = { \displaystyle \frac{1}{2}{{\cos }^{-1}}\left( \frac{1-x}{1+x} \right) } ,\,\,{ \displaystyle x\in [0,\,1] }

Lecture – 2

00:52:40

Questions Discussed:

Question 11. { \displaystyle \frac{9\pi}{8} } - { \displaystyle \frac{9}{4}{{\sin }^{-1}}\frac{1}{3} } = { \displaystyle \frac{9}{4}{{\sin }^{-1}}\frac{2\sqrt{2}}{3} }

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

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

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

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

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

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

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

Lecture – 6

00:35:01

Questions Discussed:

Question 10. Solve: { \displaystyle 3{{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}}} - { \displaystyle4{{\cos }^{-1}}\frac{1-{{x}^{2}}}{1+{{x}^{2}}}}+{ \displaystyle 2{{\tan }^{-1}}\frac{2x}{1-{{x}^{2}}}} = { \displaystyle \frac{\pi }{3} }

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

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

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

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

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

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