If 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 that 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 can write this as sec⁡θ+tan⁡θ=m\sec\theta + \tan\theta = m and label it as equation (1).

Step 2: Use the standard identity to find sec⁡θ−tan⁡θ\sec\theta - \tan\theta

Using the standard trigonometric identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, we have sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1. Factoring as the difference of two squares gives (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1.

Step 3: Express sec⁡θ−tan⁡θ\sec\theta - \tan\theta in terms of mm

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

Step 4: Add equations (1) and (2) to solve for sec⁡θ\sec\theta

Adding equation (1) and equation (2), the terms +tan⁡θ+\tan\theta and −tan⁡θ-\tan\theta cancel out, leaving 2sec⁡θ=m+1m2\sec\theta = m + \frac{1}{m}. Combining the right-hand side gives m2+1m\frac{m^2 + 1}{m}, and dividing by 2 yields sec⁡θ=m2+12m\sec\theta = \frac{m^2 + 1}{2m}.

Solve your own maths question free →