Chapter 2: Inverse Trigonometric Functions

Ans: Consider \[\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)\]

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right) $ 

$ \dfrac{\text{2cos x}}{\text{1-co}{{\text{s}}^{\text{2}}}\text{x}}\text{=2cosecx} $ 

$ \text{sin xcos x=si}{{\text{n}}^{\text{2}}}\text{x} $ 

$ \text{sin x}\left( \text{cos x-sin x} \right)\text{=0} $ 

$ \text{sin x=0,cos x-sin x=0} $ 

$ \text{x=0,}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}} $ 

Therefore, \[\text{2ta}{{\text{n}}^{\text{-1}}}\left( \text{cos x} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{2cosecx} \right)\] is satisfied for \[\text{x=0,}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\]

Ans: Consider \[\text{ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{1-x}}{\text{1+x}}\text{=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}\]

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x} \right) \right)\text{=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-ta}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{=}\dfrac{\text{3}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \text{ta}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

$ \text{x=}\dfrac{\text{1}}{\sqrt{\text{3}}} $ 

Therefore, \[\text{ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{1-x}}{\text{1+x}}\text{=}\dfrac{\text{1}}{\text{2}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x}\] is satisfied for \[\text{x=}\dfrac{\text{1}}{\sqrt{\text{3}}}\]

Ans: Assuming 

$ \text{x=tan }\!\!\beta\!\!\text{ } $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\text{x} $ 

$ \text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right) $ 

$ \text{=sin }\!\!\beta\!\!\text{ } $ 

$ \text{=}\dfrac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}} $ 

Hence, \[\text{sin}\left( \text{ta}{{\text{n}}^{\text{-1}}}\text{x} \right)\text{=}\dfrac{\text{x}}{\sqrt{\text{1+}{{\text{x}}^{\text{2}}}}}\]

Therefore, the correct option is option D

Ans: Consider

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $ 

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{-si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right) $ 

\[\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)\] -----(1)

Assuming 

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \text{1-x} \right)\text{= }\!\!\beta\!\!\text{ } $ 

$ \text{cos }\!\!\beta\!\!\text{ =1-x} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \text{1-x} \right)}^{\text{2}}}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$ \text{ }\!\!\beta\!\!\text{ =si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\text{1-x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}}\] -----(2)

Substituting Equation (2) in (1)

$\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)\text{=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$\text{-2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{=}\sqrt{\text{2x-}{{\text{x}}^{\text{2}}}} $ 

$\text{4}{{\text{x}}^{\text{2}}}\text{-4}{{\text{x}}^{\text{4}}}\text{=2x-}{{\text{x}}^{\text{2}}} $ 

$ \text{4}{{\text{x}}^{\text{4}}}\text{-5}{{\text{x}}^{\text{2}}}\text{+2x=0} $ 

$ \text{x=0,}\dfrac{\text{1}}{\text{2}}\text{,}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{17}}}{\text{4}} $ 

But considering the given options and also when \[\text{x=}\dfrac{\text{1}}{\text{2}}\], the equation \[\text{si}{{\text{n}}^{\text{-1}}}\left( \text{1-x} \right)\text{-2si}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\] doesn’t satisfy

Thus, $\text{x=0}$ is the only solution.

Hence the correct option is C