Question

Differentiate(w.r.t. x) -

tan(x-y) + tan(x+y) = 1
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[Differenciation of Implicit functions]

Answer 1

tan(x - y) + tan(x + y) = 1
differentiate with respect to x
$\frac{d[tan(x - y)+tan(x+y)]}{dx}=\frac{d.1}{dx}$


[ we know, differentiation of tanA = sec²A and differentiation of constant is equal to zero.]

so, sec²(x - y).d(x - y)/dx + sec²(x + y).d(x + y)/dx = 0
sec²(x - y).[ 1 - dy/dx] + sec²(x + y).[1 + dy/dx ]=0
[sec²(x - y) + sec²(x + y)]-dy/dx[(sec²(x - y)-sec²(x + y)] = 0
dy/dx = [sec² (x - y) + sec² (x + y)]/[sec²(x -y) - sec²(x + y)]

hence,
$\boxed{\boxed{\bold{\frac{dy}{dx}=\frac{sec^2(x-y)+sec^2(x+y)}{sec^2(x-y)-sec^2(x+y)}}}}$