Choose the correct answer from the given four option. Solution of… - Vidyadip

MCQ

Choose the correct answer from the given four option.
Solution of the differential equation $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$is:

A
$\tan\text{x}+\tan\text{y}=\text{k}$
B
$\tan\text{x}-\tan\text{y}=\text{k}$
C
$\frac{\tan\text{x}}{\tan\text{y}}=\text{k}$
✓
$\tan\text{x}.\tan\text{y}=\text{k}$

✓

Answer

Correct option: D.

$\tan\text{x}.\tan\text{y}=\text{k}$

We have, $\tan\text{y}\sec^2\text{xdx} + \tan\text{x }\sec^2\text{ydy}=0$

$\Rightarrow\tan\text{y}\sec^2\text{xdx} =- \tan\text{x }\sec^2\text{ydy}$

$\Rightarrow\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$

$\Rightarrow\int\frac{\sec^2\text{x}}{\tan\text{x}}\text{dx}=-\int\frac{\sec^2\text{y}}{\tan\text{y}}\text{dy}$

$\Rightarrow\log\tan\text{x}=-\log\tan\text{y}+\log\text{k}$

$\Rightarrow\log\tan\text{x}+\log\tan\text{y}=\log\text{k}$

$\Rightarrow\log(\tan\text{x}\tan\text{y})=\log\text{k}$

$\Rightarrow\tan\text{x}\tan\text{y}=\text{k}$

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