If cos y - x cos a - y - with cos-1- prove that dy - dx -cos 2 a - y -sin a-

Given, cos y = x cos(a+y) ⇒ x=y/cos(a+y) Differentiating both sides w.r.t. y, we get dx/dy=d/dy{cos y/cos(a+y)}=cos(a+y)( sin y) cos y( sin(a+y).1)/cos^2(a+y) ...

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Byju's Answer

Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

If cos y = x ...

Question

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

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Solution

Given, cos y = x cos(a+y) $\Rightarrow x = \frac{y}{c o s (a + y)}$

Differentiating both sides w.r.t. y, we get

$\frac{d x}{d y} = \frac{d}{d y} {\frac{c o s y}{c o s (a + y)}} = \frac{c o s (a + y) (- s i n y) - c o s y (- s i n (a + y) .1)}{c o s^{2} (a + y)} (u s i n g q u o t i e n t r u l e, \frac{d}{d y} (\frac{u}{v}) = \frac{v \frac{d}{d y} u - u \frac{d}{d y} v}{v^{2}}) = \frac{s i n (a + y) c o s y - c o s (a + y) s i n y}{c o s^{2} (a + y)} = \frac{s i n (a + y - y)}{c o s^{2} (a + y)} = \frac{s i n a}{c o s^{2} (a + y)} (∴ s i n (A - B) = s i n A c o s B - c o s A s i n B) ∴ \frac{d y}{d x} = \frac{1}{\frac{d x}{d y}} = \frac{c o s^{2} (a + y)}{s i n a} . H e n c e p r o v e d .$

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Similar questions

If $c o s y = x c o s (a + y), c o s a \neq 1$ prove that $\frac{d y}{d x} = \frac{c o s^{2} (a + y)}{s i n a}$

x cos (a + y) = cos y

then prove that

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

Q. If

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}

Q. If

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}

Q. If

x cos (a + y) = cos y,

then prove that

\frac{d y}{d x} = \frac{{c o s}^{2} (a + y)}{sin a}

. Show that

sin a \frac{d^{2} y}{d x^{2}} + sin 2 (a + y) \frac{d y}{d x} = 0

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If cos y=x cos (a+y),with cos a≠1,prove that dydx=cos2(a+y)sin a.

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