51. <math><mtable columnalign="left"><mtr><mtd><mrow><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo>(</mo><mrow><mn>1</mn><mo>−</mo><mi>x</mi></mrow><mo>)</mo></mrow><mo>−</mo><mn>2</mn><mi>s</mi><mi>i</mi><msup><mrow><mi>n</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mi>x</mi><mo>=</mo><mfrac><mrow><mi>π</mi></mrow><mrow><mn>2</mn></mrow></mfrac><mo>,</mo><mi>t</mi><mi>h</mi><mi>e</mi><mi>n</mi><mtext> </mtext><mi>x</mi><mtext> </mtext><mi>i</mi><mi>s</mi><mtext> </mtext><mi>e</mi><mi>q</mi><mi>u</mi><mi>a</mi><mi>l</mi><mtext> </mtext><mi>t</mi><mi>o</mi></mrow></mtd></mtr><mtr><mtd><mrow><mrow><mo>(</mo><mrow><mi>A</mi></mrow><mo>)</mo></mrow><mtext> </mtext><mtext> </mtext><mn>0</mn><mo>,</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mrow><mo>(</mo><mrow><mi>B</mi></mrow><mo>)</mo></mrow><mtext> </mtext><mtext> </mtext><mn>1</mn><mo>,</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mrow><mo>(</mo><mrow><mi>C</mi></mrow><mo>)</mo></mrow><mtext> </mtext><mtext> </mtext><mn>0</mn><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mtext> </mtext><mrow><mo>(</mo><mrow><mi>D</mi></mrow><mo>)</mo></mrow><mtext> </mtext><mtext> </mtext><mfrac><mrow><mn>1</mn></mrow><mrow><mn>2</mn></mrow></mfrac></mrow></mtd></mtr></mtable></math>

sin−1(1−x)−2sin−1x=π2 →(1) (M) Let x=sinθ.Then,θ=sin−1x.Putting this in qn(1) we getsin−1(1−x)−2·θ=π2⇒sin−1(1−x)=π2+2θ⇒ 1−x=sin(π2+2θ){sin(π2+x)=cosx}⇒ 1−x=cos2θ⇒ 1−x=1−2sin2θ·{cos2x=1−2sin2x}⇒1−x=1−2x2·{sinθ=x}⇒ 2x2−x=0 ⇒x(2x−1)=0⇒so, x=0 x 2x−1=0 x 2x=1→x=12.Putting x=0 in qn (1) .L.H.S =sin−1(1−0)−2sin−10=sin−1sinx2−0=π2=R.H.S.x=12q(1)L.H.S=sin−1(1−12)−2sin−112=sin−1(12)−2sin−112=sin−112−2sin−112=−sin−112=−sin−1(sinπ6)=−π6 ≠So, =0.Option (c) is correct.

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Inverse Trigonometric Functions Ncert Solutions Maths class 12th Maths Class 12th Inverse Trigonometric Functions

51. $\begin{array}{l} s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2}, t h e n x i s e q u a l t o \\ (A) 0, \frac{1}{2} (B) 1, \frac{1}{2} (C) 0 (D) \frac{1}{2} \end{array}$

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Vishal Baghel | Contributor-Level 10

5 months ago

$s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2} \to (1)$

(M) Let $x = s i n θ . T h e n, θ = s i n^{- 1} x .$

Putting this in $q^{n} (1)$ we get

$s i n^{- 1} (1 - x) - 2 \cdot θ = \frac{π}{2}$

$\Rightarrow s i n^{- 1} (1 - x) = \frac{π}{2} + 2 θ$

$\Rightarrow 1 - x = s i n (\frac{π}{2} + 2 θ) {s i n (\frac{π}{2} + x) = c o s x}$

$\Rightarrow 1 - x = c o s 2 θ$

$\Rightarrow 1 - x = 1 - 2 s i n^{2} θ \cdot {c o s 2 x = 1 - 2 s i n^{2} x}$

$\Rightarrow 1 - x = 1 - 2 x^{2} \cdot {s i n θ = x}$

$\begin{array}{l} \Rightarrow 2 x^{2} - x = 0 \\ \Rightarrow x (2 x - 1) = 0 \end{array}$

$\Rightarrow s o, x = 0 x 2 x - 1 = 0 x 2 x = 1 \to x = \frac{1}{2} .$

Putting $x = 0$ in qⁿ (1) .

L.H.S $= s i n^{- 1} (1 - 0) - 2 s i n^{- 1} 0 = s i n^{- 1} s i n \frac{x}{2} - 0 = \frac{π}{2} = R . H . S .$

$x = \frac{1}{2} q(1)$

$L.H.S= s i n^{- 1} (1 - \frac{1}{2}) - 2 s i n^{- 1} \frac{1}{2} = s i n^{- 1} (\frac{1}{2}) - 2 s i n^{- 1} \frac{1}{2}$

$= s i n^{- 1} \frac{1}{2} - 2 s i n^{- 1} \frac{1}{2}$

$= - s i n^{- 1} \frac{1}{2} = - s i n^{- 1} (s i n \frac{π}{6}) = - \frac{π}{6}$ $\neq$

So, $= 0 .$

Option (c) is correct.

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51. $\begin{array}{l} s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2}, t h e n x i s e q u a l t o \\ (A) 0, \frac{1}{2} (B) 1, \frac{1}{2} (C) 0 (D) \frac{1}{2} \end{array}$

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51. sin−1(1−x)−2sin−1x=π2,then x is equal to(A) 0,12 (B) 1,12 (C) 0 (D) 12

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51. $\begin{array}{l} s i n^{- 1} (1 - x) - 2 s i n^{- 1} x = \frac{π}{2}, t h e n x i s e q u a l t o \\ (A) 0, \frac{1}{2} (B) 1, \frac{1}{2} (C) 0 (D) \frac{1}{2} \end{array}$