Derivative of log ⁡ sec ⁡ θ + tan ⁡ θ with respect to sec ⁡ θ at θ = π 4 is _______

Step-by-step solution

Let y 1 = log ⁡ ( sec ⁡ θ + tan ⁡ θ ) ∴ d y 1 d θ = 1 sec ⁡ θ + tan ⁡ θ ⋅ ( sec ⁡ θ tan ⁡ θ + sec 2 ⁡ a ) ⇒ d y 1 d θ = sec ⁡ θ [ sec ⁡ θ + tan ⁡ θ ] [ sec ⁡ θ + tan ⁡ θ ] ⇒ d y 1 d θ = sec ⁡ θ ..... (i) Now, Let y 2 = sec ⁡ θ ∴ d y 2 d θ = sec ⁡ θ . tan ⁡ θ ..... (ii) ∴ d y 1 d y 2 = sec ⁡ θ sec ⁡ θ . tan ⁡ θ = cot ⁡ θ ∴ d y 1 d y 2 θ = π 4 = dy 1 d θ dy 2 d θ = cot ⁡ π 4 = 1