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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
MCQ
$ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$ is equal to
  • ✓
    $ \frac{\pi}{2}$
  • B
    $ \frac{\pi}{3}$
  • C
    $ \frac{\pi}{4}$
  • D
    $ \frac{\pi}{6}$
✓

Answer

Correct option: A.
$ \frac{\pi}{2}$
Given, $ \sin ^{ -1 } \frac { 3 }{ 5 } +\sin^{ -1 }\frac { 4 }{ 5 }$
$\Rightarrow \sin^{−1}\text{x}+\sin^{−1}\text{y}=\sin ^{ -1 } \Big(\text{x}\sqrt { 1-{ \text{y} }^{ 2 } } +\text{y}\sqrt { 1-\text{x}^{ 2 } }\Big)$
$ \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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