The solution of $tan x d y + tan y d x$ $=$ 0

tan x . tan y = c

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{sec}^{2} x + {sec}^{2} y = c

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sin x . sin y

= c

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cot x . cot y

= c

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Solution

The correct option is C $sin x . sin y$ $= c$
$t a n x d y + t a n y d x = 0$
$\Rightarrow \frac{- d y}{t a n y} = \frac{d x}{t a n x}$
$\Rightarrow \frac{- c o s y d y}{s i n y} = \frac{c o s x d x}{s i n x}$
$\Rightarrow \frac{- d t}{t} = \frac{d k}{k} [t = s i n y; k = c o s x]$
$\Rightarrow l o g C = l o g K t$
$\Rightarrow C = s i n x s i n y$

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Similar questions

tan x d y - tan y d x = 0

tan y d x + tan x d y = 0

The solution of D.E $y^{2} \frac{d y}{d x} + y^{2} + 1 = 0$ is

\frac{d v}{d x} = \frac{x + 2 y - 3}{2 x + y - 3} i s | x + y - 2 | = c | (x - y)^{n} |

(1 + x^{2}) \frac{d y}{d x} + 1 + y^{2} = 0

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