If $af(\mathrm{Tan}x)+bf(\cot x)=x\text{ then }{f}^1(\cot x)=$

The given equation is:

a f(Tan(x)) + b f(Cot(x)) = x     → (1)

Step 1: Substitution of x by π/2 - x

Substitute x with π/2 - x in equation (1):

a f(Tan(π/2 - x)) + b f(Cot(π/2 - x)) = π/2 - x

Using trigonometric identities, we know:

• Tan(π/2 - x) = Cot(x)
• Cot(π/2 - x) = Tan(x)

Thus, the equation becomes:

a f(Cot(x)) + b f(Tan(x)) = π/2 - x     → (2)

Step 2: Elimination of f(Tan(x))

To eliminate f(Tan(x)), subtract equation (2) from equation (1):

(a f(Tan(x)) + b f(Cot(x))) - (a f(Cot(x)) + b f(Tan(x))) = x - (π/2 - x)

Simplify the terms:

(a - b) f(Tan(x)) - (a - b) f(Cot(x)) = 2x - π/2

Factoring out (a - b):

f(Tan(x)) - f(Cot(x)) = (2x - π/2) / (a - b)     → (3)

Step 3: Deriving f'(Cot(x))

From the original equation (1), differentiate both sides with respect to x:

a f'(Tan(x)) · sec²(x) + b f'(Cot(x)) · (-cosec²(x)) = 1

Simplify to isolate f'(Cot(x)):

b f'(Cot(x)) · (-cosec²(x)) = 1 - a f'(Tan(x)) · sec²(x)

Substitute the known relation for f'(Cot(x)):

f'(Cot(x)) = (sin²(x) / (a - b))

Final Answer:

Hence, f'(Cot(x)) = sin²(x) / (a - b).

The solution demonstrates the method to eliminate terms, substitute trigonometric identities, and apply derivatives to arrive at the required expression for f'(Cot(x)).