Question
$ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$ is equal to
-
$ \frac{\pi}{2}$
-
$ \frac{\pi}{3}$
-
$ \frac{\pi}{4}$
-
$ \frac{\pi}{6}$
$ \frac{\pi}{2}$
$ \frac{\pi}{3}$
$ \frac{\pi}{4}$
$ \frac{\pi}{6}$
Solution:
Given,
$ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$$⇒\sin^{−1}\text{x}+\sin^{−1}\text{y}=\sin ^{ -1 } \Big(\text{x}\sqrt { 1-{ \text{y} }^{ 2 } } +\text{y}\sqrt { 1-\text{x}^{ 2 } }\Big)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \sqrt { 1-\left(\frac { 4 }{ 5 } \right) } +\frac { 4 }{ 5 } \sqrt { 1-\left(\frac { 3 }{ 5 } \right)^{ 2 } } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \sqrt { \frac { 25-16 }{ 25 } ) } +\frac { 4 }{ 5 } \sqrt { \frac { 25-9 }{ 25 } } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 3 }{ 5 } \times \frac { 3 }{ 5 } +\frac { 4 }{ 5 } \times \frac { 4 }{ 5 } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 16 }{ 25 } +\frac { 9 }{ 25 } \right)$
$ \Rightarrow \sin ^{ -1 } \left(\frac { 25 }{ 25 } \right)$
$ \Rightarrow \sin ^{ -1 } (1)$
$ \Rightarrow \cfrac { \pi }{ 2 }$
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| $\text{X}:$ | $2$ | $3$ | $4$ | $5$ |
| $\text{P}(\text{X}):$ | $\frac{5}{\text{k}}$ | $\frac{7}{\text{k}}$ | $\frac{9}{\text{k}}$ | $\frac{11}{\text{k}}$ |
The value of k is:
$(1)$ $y=\log _0\left(\frac{1+\sqrt{1-x^2}}{x}\right)-\sqrt{1-x^2}$
$(2)$ $x y^{\prime}-\sqrt{1-x^2}=0$
$(3)$ $y=-\log _0\left(\frac{1+\sqrt{1-x^2}}{x}\right)+\sqrt{1-x^2}$
$(4)$ $x y^{\prime}+\sqrt{1-x^2}=0$