Question

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Answer 1

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here's the solution

Answer 2

Given : Secθ - Tanθ = x ---------- [1]

Multiplying Both Sides with (Secθ + Tanθ)

⇒ (Secθ - Tanθ)(Secθ + Tanθ) = (x)(Secθ + Tanθ)

⇒ Sec²θ - Tan²θ = (x)(Secθ + Tanθ)

We know that : Sec²θ - Tan²θ = 1

⇒ (x)(Secθ + Tanθ) = 1

⇒ (Secθ + Tanθ) = $\frac{1}{x}$ ---------------- [2]

Now Adding Both Equation [1] and Equation [2], We get :

⇒ (Secθ - Tanθ) + (Secθ + Tanθ) = $(x + \frac{1}{x})$

⇒ 2Secθ = $(x + \frac{1}{x})$

⇒ Secθ = $\frac{1}{2}(x + \frac{1}{x})$

Subtracting Equation [1] from Equation [2], We get :

⇒ (Secθ + Tanθ) - (Secθ - Tanθ) = $(\frac{1}{x} - x)$

⇒ 2Tanθ = $(\frac{1}{x} - x)$

⇒ Tanθ = $\frac{1}{2}( \frac{1}{x} - x)$