Subjective Type

Prove: $$\displaystyle \frac { 9\pi }{ 8 } -\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 1 }{ 3 }=\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } $$

Solution

$$\displaystyle L.H.S.=\frac { 9\pi }{ 8 } -\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 1 }{ 3 } $$

$$\displaystyle =\frac { 9 }{ 4 } \left( \frac { \pi }{ 2 } -{ \sin }^{ -1 }\frac { 1 }{ 3 } \right) $$

$$\displaystyle=\frac { 9 }{ 4 } \left( { \cos }^{ -1 }\frac { 1 }{ 3 } \right) .....(1)$$

Now, let $$\displaystyle { \cos }^{ -1 }\frac{ 1 }{ 3 }=x $$.

Then, $$\displaystyle \cos { x } =\frac { 1 }{ 3 } \Rightarrow \sin { x } =\sqrt { 1-{ \left( \frac { 1 }{ 3 } \right) }^{ 2 } } =\frac { 2\sqrt { 2 } }{ 3 } $$

$$\displaystyle \therefore x={ \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } \Rightarrow {
\cos }^{ -1 }\frac { 1 }{ 3 } ={ \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } $$

$$\displaystyle \therefore \text{L.H.S.}=\frac { 9 }{ 4 } { \sin }^{ -1 }\frac { 2\sqrt { 2 } }{ 3 } =\text{R.H.S.}$$

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