If-cosec-1-x-cot-1-y-show-that-d-2-y-dx-2-1-y-3-0-

Question Number 117482 by Don08q last updated on 12/Oct/20

If {cosec}^{- 1} x = \cot^{- 1} y, show that

\frac{d^{2} y}{{d x}^{2}} + \frac{1}{y^{3}} = 0

Answered by TANMAY PANACEA last updated on 12/Oct/20

y = c o t ({c o s e c}^{- 1} x)

θ = {c o s e c}^{- 1} x \to c o s e c θ = x

x^{2} = 1 + {c o t}^{2} θ \to c o t θ = \sqrt{x^{2} - 1}

y = \sqrt{x^{2} - 1}

\frac{d y}{d x} = \frac{1}{2 \sqrt{x^{2} - 1}} \times 2 x = \frac{x}{\sqrt{x^{2} - 1}}

\frac{d^{2} y}{{d x}^{2}} = \frac{\sqrt{x^{2} - 1} \times 1 - x \times \frac{1}{2 \sqrt{x^{2} - 1}} \times 2 x}{x^{2} - 1}

= \frac{x^{2} - 1 - x^{2}}{{(x^{2} - 1)}^{\frac{3}{2}}} = \frac{- 1}{\boldsymbol y^{3}} \boldsymbol p r o v e d

Commented by $@y@m last updated on 12/Oct/20

Nice. I just want to add that there is no use of writing first line.

Commented by TANMAY PANACEA last updated on 13/Oct/20

t h a n k y o u s i r

Commented by Don08q last updated on 13/Oct/20

T h a n k y o u S i r

Answered by $@y@m last updated on 13/Oct/20

L e t {cosec}^{- 1} x = \cot^{- 1} y = θ

x = cosec θ ∣ y = \cot θ

\frac{d x}{d θ} = - cosec θ \cot θ ∣ \frac{d y}{d θ} = - cosec^{2} θ

\frac{d y}{d x} = \frac{cosec θ}{\cot θ}

\frac{d y}{d x} = \sec θ \dots . . (1)

\frac{d^{2} y}{{d x}^{2}} = \frac{d}{d x} (\sec θ) = \frac{d}{d θ} (\sec θ) \times \frac{d θ}{d x}

\frac{d^{2} y}{{d x}^{2}} = \sec θ \tan θ \times \frac{1}{- cosec θ \cot θ}

\frac{d^{2} y}{{d x}^{2}} = \frac{1}{- \cot^{3} θ} = - \frac{1}{y^{3}}

\frac{d^{2} y}{{d x}^{2}} + \frac{1}{y^{3}} = 0

Commented by Don08q last updated on 13/Oct/20

T h a n k y o u S i r

T h a n k y o u S i r

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