If x=secθ−cosθ,y=sec10θ−cos10θ and (x2+4

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If $x = s e c θ - c o s θ, y = s e c^{10} θ - c o s^{10} θ$ and $(x^{2} + 4) {(\frac{d y}{d x})}^{2} = k (y^{2} + 4)$ , then k is equal to

Options:

A .

\frac{1}{100}

B .

1

C .

10

D .

100

Answer: Option D
:
D

∵ x^{2} + 4 = (s e c θ - c o s θ)^{2} + 4 = (s e c θ + c o s θ)^{2} \dots (i) S i m i l a r l y, y^{2} + 4 = (s e c^{10} θ + c o s^{10} θ)^{2} \dots (i i) N o w, \frac{d x}{d θ} = s e c θ t a n θ + s i n θ = t a n θ (s e c θ + c o s θ) a n d \frac{d y}{d θ} = 10 s e c^{9} θ s e c θ t a n θ - 10 c o s^{9} θ (- s i n θ) = 10 t a n θ (s e c^{10} θ - c o s^{10} θ) \Rightarrow \frac{d y}{d x} = \frac{(\frac{d y}{d θ})}{(\frac{d x}{d θ})} = \frac{10 t a n θ (s e c^{10} θ + c o s^{10} θ)}{t a n θ (s e c θ + c o s θ)} ∴ {(\frac{d y}{d x})}^{2} = 100 \frac{(s e c^{10} θ + c o s^{10} θ)}{(s e c θ + c o s θ)^{2}} = \frac{100 (y^{2} + 4)}{(x^{2} + 4)} o r (x^{2} + 4) {(\frac{d y}{d x})}^{2} = 100 (y^{2} + 4)

On comparing with the expression given we get k = 100

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