If y -sin -1 x 2-cos -1 x 2- then 1- x 2 d2 y-d x 2- x dy - dx -A- 2B- -2C- 4D- -4

The correct option is C 4y_1=2 sin^ 1x/√(1 x^2) 2 cos^ 1x/√(1 x^2) (√(1 x^2))y_1 = 2 sin^ 1x 2 cos^ 1x Differentiating w.r.t x again we get (1 x^2)d^2 y/dx^ ...

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Byju's Answer

Standard XI

Mathematics

Solving Simultaneous Trigonometric Equations

If y =sin -1 ...

Question

If $y = (s i n^{- 1} x)^{2} + (c o s^{- 1} x)^{2}$ , then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} =$

-4

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Solution

The correct option is B 4
$y_{1} = \frac{2 s i n^{- 1} x}{\sqrt{1 - x^{2}}} - \frac{2 c o s^{- 1} x}{\sqrt{1 - x^{2}}}$

$(\sqrt{1 - x^{2}}) y_{1} = 2 s i n^{- 1} x - 2 c o s^{- 1} x$

Differentiating w.r.t x again we get

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 4$

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Similar questions

Q. If

y = (s i n^{- 1} x)^{2} + (c o s^{- 1} x)^{2}, t h e n (1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} =

Q. if

y = ({sin}^{- 1} x)^{2} + k {sin}^{- 1} x

, then

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x}

Q. If

y = ({sin}^{- 1} x)^{2}

, then show that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 2

Q. If

y = ({sin}^{- 1} x)^{2}

, then prove that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 2

Q. If

y = {({sin}^{- 1} x)}^{2} + c,

then prove that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 2

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If y=(sin−1x)2+(cos−1x)2, then (1−x2)d2ydx2−xdydx=

The correct option is B 4y1=2 sin−1x√1−x2−2 cos−1x√1−x2 (√1−x2)y1=2 sin−1x−2 cos−1x Differentiating w.r.t x again we get (1−x2)d2ydx2−xdydx=4

If $y = (s i n^{- 1} x)^{2} + (c o s^{- 1} x)^{2}$ , then $(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} =$

The correct option is B 4
$y_{1} = \frac{2 s i n^{- 1} x}{\sqrt{1 - x^{2}}} - \frac{2 c o s^{- 1} x}{\sqrt{1 - x^{2}}}$

$(\sqrt{1 - x^{2}}) y_{1} = 2 s i n^{- 1} x - 2 c o s^{- 1} x$

Differentiating w.r.t x again we get

$(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = 4$