Inverse Trigonometric Functions Part - 2

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] }

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} }

Questions Discussed:

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

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

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

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

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

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

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

Questions Discussed:

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

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

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

Solve:

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

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

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

Questions Discussed:

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

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

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

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

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

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

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 } .