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

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

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Byju's Answer

Standard XII

Physics

Basic Differentiation Rule

If x+cos q= q...

Question

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

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Solution

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

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If x+cos q = sec q, y + cos8 q=sec8 q then (x2+4y2+4)(dydx)2=

The correct option is C 64dydx=dydqdxdq=8sec8q tan q+8 cos7 q sin qsec q tan q + sin q =8 tan q(sec8 q+cos8 q)tan q (sec q+cos q) (dydx)2=64(sec8q+cos8q)2(secq+cosq)2 =64[(sec8q−cos8q)2+4][(secq−cosq)2+4] =64(y2+4)x2+4 (x2+4y2+4)(dydx)2=64

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