35) tan⁡θ+sec⁡θ=m\tan\theta + \sec\theta = m, prove that sec⁡θ=m2+12m\sec\theta = \frac{m^2 + 1}{2m}

Answer: Hence proved, sec⁡θ=m2+12m\sec\theta = \frac{m^2 + 1}{2m}.

Step-by-step solution

Step 1: State the given equation

We are given that tan⁡θ+sec⁡θ=m\tan \theta + \sec \theta = m. We write this as sec⁡θ+tan⁡θ=m\sec \theta + \tan \theta = m and label it as Equation (1).

Step 2: Use the trigonometric identity

We recall the fundamental trigonometric identity sec⁡2θ−tan⁡2θ=1\sec^2 \theta - \tan^2 \theta = 1. Using the algebraic identity a2−b2=(a−b)(a+b)a^2 - b^2 = (a - b)(a + b), we factor the expression into (sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=1(\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1.

Step 3: Find the expression for sec⁡θ−tan⁡θ\sec \theta - \tan \theta

Substitute sec⁡θ+tan⁡θ=m\sec \theta + \tan \theta = m into the factored identity. Dividing both sides by mm, we obtain sec⁡θ−tan⁡θ=1m\sec \theta - \tan \theta = \frac{1}{m}, which we label as Equation (2).

Step 4: Add Equation (1) and Equation (2)

We add Equation (1) and Equation (2). The terms tan⁡θ\tan \theta and −tan⁡θ-\tan \theta cancel each other out, leaving 2sec⁡θ=m+1m=m2+1m2\sec \theta = m + \frac{1}{m} = \frac{m^2 + 1}{m}.

Step 5: Solve for sec⁡θ\sec \theta

Dividing both sides of the equation by 2 yields sec⁡θ=m2+12m\sec \theta = \frac{m^2 + 1}{2m}, which completes the proof.

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