Considering the principal values of the inverse trigonometric functions, , , is equal to [2025]

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Q.
Considering the principal values of the inverse trigonometric functions, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$ , $- \frac{1}{2} < x < \frac{1}{\sqrt{2}}$ , is equal to [2025]

1 $\frac{- 5 π}{6} - \sin^{- 1} x$

2 $\frac{π}{4} + \sin^{- 1} x$

3 $\frac{5 π}{6} - \sin^{- 1} x$

4 $\frac{π}{6} + \sin^{- 1} x$

Ans.
(4)

Let $\sin^{- 1} x = θ, \frac{- π}{4} < θ < \frac{π}{4}$

$\Rightarrow x = \sin θ$

Now, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$

$= \sin^{- 1} (\sin θ \cos \frac{π}{6} + \cos θ \sin \frac{π}{6})$

$= \sin^{- 1} (\sin (θ + \frac{π}{6})) = θ + \frac{π}{6} = \sin^{- 1} x + \frac{π}{6}$

Solution

Q.
Considering the principal values of the inverse trigonometric functions, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$ , $- \frac{1}{2} < x < \frac{1}{\sqrt{2}}$ , is equal to [2025]

1 $\frac{- 5 π}{6} - \sin^{- 1} x$

2 $\frac{π}{4} + \sin^{- 1} x$

3 $\frac{5 π}{6} - \sin^{- 1} x$

4 $\frac{π}{6} + \sin^{- 1} x$

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Solution

Q. Considering the principal values of the inverse trigonometric functions, sin–1(32x+121–x2), –12<x<12, is equal to [2025]

1 –5π6–sin–1x

2 π4+sin–1x

3 5π6–sin–1x

4 π6+sin–1x

Ans. (4) Let sin–1x=θ,–π4<θ<π4 ⇒ x=sinθ Now, sin–1(32x+121–x2) =sin–1(sinθ cosπ6+cosθ sinπ6) =sin–1(sin(θ+π6))=θ+π6=sin–1x+π6

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Q.
Considering the principal values of the inverse trigonometric functions, $\sin^{- 1} (\frac{\sqrt{3}}{2} x + \frac{1}{2} \sqrt{1 - x^{2}})$ , $- \frac{1}{2} < x < \frac{1}{\sqrt{2}}$ , is equal to [2025]

1 $\frac{- 5 π}{6} - \sin^{- 1} x$

2 $\frac{π}{4} + \sin^{- 1} x$

3 $\frac{5 π}{6} - \sin^{- 1} x$

4 $\frac{π}{6} + \sin^{- 1} x$