If x-cos q-secq- y-cos 8 q- 8 q then x2-4-y2-4 d x

The correct option is B 64dy/dx=dy/dq/dx/dq=8 sec^8q tan q+8 cos^7qsin q/sec q tan q+sin q=8 tan q(sec^8q+cos^8q)/tan q(sec q+cos q) ( dy/dx )^2=64(sec^8θ+cos ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

If x+cos q=se...

Question

If $x + c o s q = s e c q, y + c o s^{8} q = s e c^{8} q t h e n \frac{x^{2} + 4}{y^{2} + 4} \frac{d y}{d x}$

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Solution

The correct option is C 64
$\frac{d y}{d x} = \frac{\frac{d y}{d q}}{\frac{d x}{d q}} = \frac{8 s e c^{8} q t a n q + 8 c o s^{7} q s i n q}{s e c q t a n q + s i n q} = \frac{8 t a n q (s e c^{8} q + c o s^{8} q)}{t a n q (s e c q + c o s q)} {(\frac{d y}{d x})}^{2} = \frac{64 (s e c^{8} θ + c o s^{8} θ)}{(s e c θ + c o s θ)^{2}} = \frac{64 [(s e c^{8} θ - c o s^{8} θ)^{2} + 4]}{[(s e c θ - c o s θ)^{2} + 4]} = \frac{64 (y^{2} + 4)}{(x^{2} + 4)} (\frac{x^{2} + 4}{y^{2} + 4}) {(\frac{d y}{d x})}^{2} = 64$

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If x+cosq=secq,y+cos8q=sec8q then x2+4y2+4dydx

The correct option is C 64dydx=dydqdxdq=8sec8qtanq+8cos7qsinqsecqtanq+sinq=8tanq(sec8q+cos8q)tanq(secq+cosq)(dydx)2=64(sec8θ+cos8θ)(secθ+cosθ)2=64[(sec8θ−cos8θ)2+4][(secθ−cosθ)2+4]=64(y2+4)(x2+4)(x2+4y2+4)(dydx)2=64

If $x + c o s q = s e c q, y + c o s^{8} q = s e c^{8} q t h e n \frac{x^{2} + 4}{y^{2} + 4} \frac{d y}{d x}$