Chapter 2: Inverse Trigonometric Functions

Ans: Consider $\text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}}$

$ \text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} $ 

$ \text{=cos}\left( \text{2 }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

$ \text{=cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

Further solving,

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right) $ 

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right) $ 

$\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $ 

Therefore, $\text{co}{{\text{s}}^{\text{-1}}}\left( \text{cos}\dfrac{\text{13 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

Ans: Consider $\text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}}$

$ \text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} $

$ \text{=tan}\left( \text{ }\!\!\pi\!\!\text{ +}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

$ \text{=tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

Further solving,

$ \text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right) $

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} \right) \right) $

$ \text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}} $

Therefore, $\text{ta}{{\text{n}}^{\text{-1}}}\left( \text{tan}\dfrac{\text{7 }\!\!\pi\!\!\text{ }}{\text{6}} \right)\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{6}}$

Ans: Assuming \[\text{2si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }\] ----(1)

$ \text{sin}\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\dfrac{\text{3}}{\text{5}} $ 

$ \text{cos}\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\sqrt{\text{1-}{{\left( \dfrac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

 $ \text{cos}\dfrac{\text{ }\!\!\alpha\!\!\text{ }}{\text{2}}\text{=}\dfrac{\text{4}}{\text{5}} $ 

Hence,

$ \text{sin }\!\!\alpha\!\!\text{ =2}\left( \dfrac{\text{3}}{\text{5}} \right)\left( \dfrac{\text{4}}{\text{5}} \right) $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\dfrac{\text{24}}{\text{25}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{24}}{\text{25}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{7}}{\text{25}} $ 

Therefore, 

\[\text{tan }\!\!\alpha\!\!\text{ =}\dfrac{\text{24}}{\text{7}}\]

\[\text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{24}}{\text{7}} \right)\]

From Equation (1), it is proved that 

\[\text{2si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{24}}{\text{7}} \right)\]

Ans: Assuming \[\text{si}{{\text{n}}^{\text{-1}}}\dfrac{\text{8}}{\text{17}}\text{= }\!\!\alpha\!\!\text{ }\] 

$\text{sin }\!\!\alpha\!\!\text{ =}\dfrac{\text{8}}{\text{17}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{8}}{\text{17}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{15}}{\text{17}} $ 

Hence,

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\dfrac{\text{8}}{\text{15}} $ 

$ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{15}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{17}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{15}} \right)\] ----(1) 

Assuming that \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{sin }\!\!\beta\!\!\text{ =}\dfrac{\text{3}}{\text{5}} $ 

$ \text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

$\text{cos }\!\!\beta\!\!\text{ =}\dfrac{\text{4}}{\text{5}} $ 

Therefore, 

$ \text{tan }\!\!\beta\!\!\text{ =}\dfrac{\text{3}}{\text{4}} $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right)\] ----(2) 

Consider 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{17}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{5}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{8}}{\text{15}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\dfrac{\text{8}}{\text{15}}\text{+}\dfrac{\text{3}}{\text{4}}}{\text{1-}\left( \dfrac{\text{8}}{\text{15}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{4}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{77}}{\text{36}} \right) $ 

Hence, it is proved that \[\text{si}{{\text{n}}^{\text{-1}}}\dfrac{\text{8}}{\text{17}}\text{+si}{{\text{n}}^{\text{-1}}}\dfrac{\text{3}}{\text{5}}\text{=ta}{{\text{n}}^{\text{-1}}}\dfrac{\text{77}}{\text{36}}\]

Ans: Assuming \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{4}}{\text{5}}\text{= }\!\!\alpha\!\!\text{ }\] 

$\text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{4}}{\text{5}} $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{4}}{\text{5}} \right)}^{\text{2}}}} $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\dfrac{\text{3}}{\text{5}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\dfrac{\text{3}}{\text{4}} $ 

 $ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{4}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right)\] ----(1) 

Assuming that \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{cos }\!\!\beta\!\!\text{ =}\dfrac{\text{12}}{\text{13}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{12}}{\text{13}} \right)}^{\text{2}}}} $ 

 $ \text{sin }\!\!\beta\!\!\text{ =}\dfrac{\text{5}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\beta\!\!\text{ =}\dfrac{\text{5}}{\text{12}} $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{5}}{\text{12}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{5}}{\text{12}} \right)\] ----(2) 

Consider 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{12}}{\text{13}} \right)\]

From Equations (1) and (2),

$\text{co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{4}}{\text{5}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{12}}{\text{13}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3}}{\text{4}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{5}}{\text{12}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\dfrac{\text{3}}{\text{4}}\text{+}\dfrac{\text{5}}{\text{12}}}{\text{1-}\left( \dfrac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{56}}{\text{33}} \right) $ 

$ \text{=co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{33}}{\text{65}} \right) $ 

Hence, it is proved that \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{4}}{\text{5}}\text{+co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{=co}{{\text{s}}^{\text{-1}}}\dfrac{\text{33}}{\text{65}}\]

Ans: Assuming \[\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{12}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }\] 

$ \text{cos }\!\!\alpha\!\!\text{ =}\dfrac{\text{12}}{\text{13}} $ 

$ \text{sin }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \dfrac{\text{12}}{\text{13}} \right)}^{\text{2}}}} $ 

$\text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}} $ 

$ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\] ----(1) 

Assuming that \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$\text{sin }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

$ \text{cos }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

Therefore, 

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{4}} $ 

$ \text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right)\] ----(2) 

Consider 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{12}}{\text{13}} \right)\text{+si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{3}}{\text{4}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{3}}{\text{4}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{3}}{\text{4}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{33}} \right) $ 

$ \text{=si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{56}}{\text{65}} \right) $ 

Hence, it is proved that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{12}}{\text{13}}\text{+si}{{\text{n}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{56}}{\text{65}}\]

Ans: Assuming \[\text{si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{= }\!\!\alpha\!\!\text{ }\] 

$ \text{sin }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{13}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{5}}{\text{13}} \right)}^{\text{2}}}} $ 

$ \text{cos }\!\!\alpha\!\!\text{ =}\frac{\text{12}}{\text{13}} $ 

Therefore, 

$ \text{tan }\!\!\alpha\!\!\text{ =}\frac{\text{5}}{\text{12}} $ 

$ \text{ }\!\!\alpha\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right) $ 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\] ----(1) 

Assuming that \[\text{co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\text{= }\!\!\beta\!\!\text{ }\]

$ \text{cos }\!\!\beta\!\!\text{ =}\frac{\text{3}}{\text{5}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\sqrt{\text{1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}} $ 

$ \text{sin }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{5}} $ 

Therefore, 

$\text{tan }\!\!\beta\!\!\text{ =}\frac{\text{4}}{\text{3}} $ 

$\text{ }\!\!\beta\!\!\text{ =ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right) $ 

\[\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right)\] ----(2) 

Consider 

\[\text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right)\]

From Equations (1) and (2),

$ \text{si}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{13}} \right)\text{+co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{3}}{\text{5}} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{5}}{\text{12}} \right)\text{+ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{4}}{\text{3}} \right) $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\frac{\text{4}}{\text{3}}\text{+}\frac{\text{5}}{\text{12}}}{\text{1-}\left( \frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\frac{\text{5}}{\text{12}} \right)} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \frac{\text{63}}{\text{16}} \right) $ 

Hence, it is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\frac{\text{63}}{\text{16}}\text{=si}{{\text{n}}^{\text{-1}}}\frac{\text{5}}{\text{13}}\text{+co}{{\text{s}}^{\text{-1}}}\frac{\text{3}}{\text{5}}\]

Ans: Assuming \[\text{x=ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }\] 

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}} $ 

 $ \text{=ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\text{tan }\!\!\alpha\!\!\text{ } $ 

$\text{= }\!\!\alpha\!\!\text{ } $ 

\[\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{= }\!\!\alpha\!\!\text{ }\]----(1)

Consider 

$\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right) $ 

$ \text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }}{\text{1+ta}{{\text{n}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }} \right) $ 

$ \text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{cos2 }\!\!\alpha\!\!\text{ } $ 

$ \text{= }\!\!\alpha\!\!\text{ } $ 

\[\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\text{= }\!\!\alpha\!\!\text{ }\] -----(2)

From Equations (1) and (2),

It is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}}\text{=}\frac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{1-x}}{\text{1+x}} \right)\]

Ans: We know that 

$\sqrt{\text{1+sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} $ 

$ \sqrt{\text{1-sin x}}\text{=cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} $ 

Consider

$\text{co}{{\text{t}}^{\text{-1}}}\left( \frac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right) $ 

$ \text{=co}{{\text{t}}^{\text{-1}}}\left( \frac{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{+}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)}{\left( \text{cos}\frac{\text{x}}{\text{2}}\text{+sin}\frac{\text{x}}{\text{2}} \right)\text{-}\left( \text{cos}\frac{\text{x}}{\text{2}}\text{-sin}\frac{\text{x}}{\text{2}} \right)} \right) $ 

$ \text{co}{{\text{t}}^{\text{-1}}}\left( \text{cot}\dfrac{\text{x}}{\text{2}} \right) $ 

$ \text{=}\dfrac{\text{x}}{\text{2}} $ 

It is proved that \[\text{co}{{\text{t}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{1+sin x}}\text{+}\sqrt{\text{1-sin x}}}{\sqrt{\text{1+sin x}}\text{-}\sqrt{\text{1-sin x}}} \right)\text{=}\dfrac{\text{x}}{\text{2}}\]

Ans: Assuming

$\text{x=cos2 }\!\!\beta\!\!\text{ } $ 

$ \sqrt{\text{1+x}} $ 

$ \text{=}\sqrt{\text{1+cos2 }\!\!\beta\!\!\text{ }} $ 

$ \text{=}\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ } $ 

$ \sqrt{\text{1-x}} $ 

$ \text{=}\sqrt{\text{1-cos2 }\!\!\beta\!\!\text{ }} $ 

$ \text{=}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ } $

Consider

$\text{ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right) $ 

$\text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ -}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }}{\sqrt{\text{2}}\text{cos }\!\!\beta\!\!\text{ +}\sqrt{\text{2}}\text{sin }\!\!\beta\!\!\text{ }} \right) $ 

$ \text{=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{1-tan }\!\!\beta\!\!\text{ }}{\text{1+tan }\!\!\beta\!\!\text{ }} \right) $ 

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

$\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{- }\!\!\beta\!\!\text{ } $ 

$ \text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x} $ 

It is proved that \[\text{ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\sqrt{\text{1+x}}\text{-}\sqrt{\text{1-x}}}{\sqrt{\text{1+x}}\text{+}\sqrt{\text{1-x}}} \right)\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{x}\]