If x-sin t- y-t cos t- d y-d x-A- 1- t tantB- 1-t tantC- 1 -tantD- tant - t tant

The correct option is B 1 t tantx=f(t)=sin t dx/dt=f'(t)=cos t y=g(t)=t cos t dy/dt=g'(t)=cos t t sin t dy/dx=g'(t)/f'(t) =cos t t sin t/cos t =1 t tan t ...

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Standard XI

Mathematics

Algebra of Derivatives

If x=sin t, y...

Question

If $x = s i n t, y = t c o s t, \frac{d y}{d x} =$ .

1 + t tant

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1 - t tant

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Solution

The correct option is B 1 - t tant
$x = f (t) = s i n t \frac{d x}{d t} = f^{'} (t) = c o s t$
$y = g (t) = t c o s t \frac{d y}{d t} = g^{'} (t) = c o s t - t s i n t$
$\frac{d y}{d x} = \frac{g^{'} (t)}{f^{'} (t)} = \frac{c o s t - t s i n t}{c o s t} = 1 - t t a n t$

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If x=sin t, y=tcos t, dydx=.

The correct option is B 1 - t tantx=f(t)=sin tdxdt=f′(t)=cos t y=g(t)=t cos tdydt=g′(t)=cos t−t sin t dydx=g′(t)f′(t)=cos t−t sin tcos t=1−t tan t

If $x = s i n t, y = t c o s t, \frac{d y}{d x} =$ .

The correct option is B 1 - t tant
$x = f (t) = s i n t \frac{d x}{d t} = f^{'} (t) = c o s t$
$y = g (t) = t c o s t \frac{d y}{d t} = g^{'} (t) = c o s t - t s i n t$
$\frac{d y}{d x} = \frac{g^{'} (t)}{f^{'} (t)} = \frac{c o s t - t s i n t}{c o s t} = 1 - t t a n t$