Then, equation (1) reduces to ⇒sina+y ydydx=cos2a+y⇒dydx=cos2a+ysina Hence, proved. ...

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Cramer's Rule

If with prov...

Question

If with prove that

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Solution

Then, equation (1) reduces to

$\Rightarrow \sin (a + y - y) \frac{d y}{d x} = \cos^{2} (a + y) \Rightarrow \frac{d y}{d x} = \frac{\cos^{2} (a + y)}{\sin a}$

Hence, proved.

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Similar questions

If, prove that

\cos y = x \cos (a + y), with \cos a \neq \pm 1, prove that \frac{d y}{d x} = \frac{\cos^{2} (a + y)}{\sin a}

\cos y = x \cos (a + y), with \cos a \neq \pm 1, prove that \frac{d y}{d x} = \frac{\cos^{2} (a + y)}{\sin a}

cos y = x cos (a + y)

cos a \neq \pm 1

\frac{d y}{d x} = \frac{{cos}^{2} (a + y)}{sin a}

$I f c o s y = x c o s (a + y), w i t h c o s a \neq 1, p r o v e t h a t \frac{d y}{d x} = \frac{c o s^{2} (a + y)}{s i n a} .$

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