Question

sec^2xtanydx+sec^2ytanxdy=0.​

Answer 1

EXPLANATION.

$\implies sec^{2}x \ tany dx \ + sec^{2} y \ tanx dx = 0.$

As we know that,

Divide L.H.S & R.H.S with tan(y).tan(x), we get.

$\implies \dfrac{sec^{2} x \ tan (y )dx \ + sec^{2} y \ tan(x )dy}{tan(y). tan(x)} \ = \dfrac{0}{tan(y).tan(x))}$

$\implies \dfrac{sec^{2}x \ tan (y)dx }{tan(y).tan(x)} \ + \dfrac{sec^{2}y \ tan(x)dy }{tan(y).tan(x)} \ = 0$

$\implies \dfrac{sec^{2}x }{tan(x)} dx \ + \dfrac{sec^{2}y }{tan(y)} dy \ = 0$

Integrate both sides, we get.

$\implies \displaystyle \int \dfrac{sec^{2}x }{tan(x)} dx \ + \ \int \dfrac{sec^{2}y }{tan(y)} dy \ = 0$

As we know that,

By using substitution method, we get.

Let we assume that,

⇒ tan(x) = t.

Differentiate w.r.t x, we get.

⇒ sec²xdx = dt.

⇒ tan(y) = z.

Differentiate w.r.t y, we get.

⇒ tan(y) = z.

⇒ sec²ydy = dz.

Put the value in the equation, we get.

$\implies \displaystyle \int \dfrac{dt}{t} \ + \int \dfrac{dz}{z} \ = 0$

$\implies log|t| \ + \ log|z| \ = log(C)$

Put the value of t = tan(x)  and  z = tan(y) in the equation, we get.

$\implies log|tan(x)| \ + log|tan(y)| \ = log|C|$

$\implies tan(x).tan(y) \ = C$