Q84 / 150 · MHT CET 2026 (15 April Shift 2) · Mathematics

The solution of $\dfrac{\text{d}y}{\text{d}x} = \sin(x + y) + \cos(x + y)$ is

  • A
    $\log\left[1 + \tan\left(\dfrac{x+y}{2}\right)\right] = y + c$, where c is the constant of integration
  • B
    $\log\left[1 - \tan\left(\dfrac{x+y}{2}\right)\right] = y + c$, where c is the constant of integration
  • Correct answer
    C
    $\log\left[1 + \tan\left(\dfrac{x+y}{2}\right)\right] = x + c$, where c is the constant of integration
  • D
    $\log\left[1 - \tan\left(\dfrac{x+y}{2}\right)\right] = x + c$, where c is the constant of integration

Solution

Let $x + y = v$

Differentiating with respect to $x$, we get $1 + \dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}v}{\text{d}x}$

$\dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}v}{\text{d}x} - 1$

Substituting in the given differential equation:

$\dfrac{\text{d}v}{\text{d}x} - 1 = \sin v + \cos v$

$\dfrac{\text{d}v}{\text{d}x} = 1 + \sin v + \cos v$

$\dfrac{\text{d}v}{1 + \sin v + \cos v} = \text{d}x$

Using the half-angle formulas $\sin v = \dfrac{2\tan(v/2)}{1+\tan^2(v/2)}$ and $\cos v = \dfrac{1-\tan^2(v/2)}{1+\tan^2(v/2)}$:

$\dfrac{\text{d}v}{1 + \dfrac{2\tan(v/2)}{1+\tan^2(v/2)} + \dfrac{1-\tan^2(v/2)}{1+\tan^2(v/2)}} = \text{d}x$

$\dfrac{(1+\tan^2(v/2))\text{d}v}{1+\tan^2(v/2) + 2\tan(v/2) + 1-\tan^2(v/2)} = \text{d}x$

$\dfrac{\sec^2(v/2)\text{d}v}{2(1+\tan(v/2))} = \text{d}x$

Integrating both sides:

$\int \dfrac{\sec^2(v/2)}{2(1+\tan(v/2))} \text{d}v = \int \text{d}x$

Let $1 + \tan(v/2) = t$, then $\dfrac{1}{2}\sec^2(v/2)\text{d}v = \text{d}t$

$\int \dfrac{\text{d}t}{t} = \int \text{d}x$

$\log t = x + c$

$\log\left[1 + \tan\left(\dfrac{v}{2}\right)\right] = x + c$

Substituting $v = x + y$:

$\log\left[1 + \tan\left(\dfrac{x+y}{2}\right)\right] = x + c$

Answer: $\log\left[1 + \tan\left(\dfrac{x+y}{2}\right)\right] = x + c$, where c is the constant of integration

Hence, the correct answer is option: 'C'.

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