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] }