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General Solution of Trigonometric Equation

If y=tan -1 x...

Question

If $y = (t a n^{- 1} x)^{2}, t h e n (x^{2} + 1)^{2} y_{2} + 2 x (x^{2} + 1) y_{1}$ is equal to

A

4

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B

0

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C

2

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D

1

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Solution

The correct option is C
2

$y = (t a n^{- 1} x)^{2} y_{1} = \frac{2 t a n^{- 1} x}{1 + x^{2}} y_{2} = \frac{2 (1 - 2 x t a n^{- 1} x)}{(1 + x^{2})^{2}}$

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Similar questions

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

(x^{2} + 1)^{2} y_{2} + 2 x (x^{2} + 1) y_{1}

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

(x^{2} + 1)^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2

y = (t a n^{- 1} x)^{2}

{(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2

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

{(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2

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

{(x^{2} + 1)}^{2} y_{2} + 2 x (x^{2} + 1) y_{1} = 2

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General Solution of Trigonometric Equation

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