If tan θ + sec θ = m, then prove that sec θ = (m2 + 1)/2m .

Q) If tan θ + sec θ = m, then prove that sec θ = $\frac{m^2 + 1}{2m}$ .

Ans: We are given: tan θ + sec θ = m ………… (i)

Next, we calculate value of tan θ + sec θ

To do that, we multiply and divide (tan θ + sec θ) by (tan θ – sec θ)

Hence, (tan θ + sec θ) $\frac{(\tan \theta - \sec \theta)} {(\tan \theta - \sec \theta)}$ = m

= $\frac{(\tan ^2 \theta - \sec ^2 \theta)} {(\tan \theta - \sec \theta)}$ = m

We know that 1 + tan² θ = sec² θ

∴ tan² θ – sec² θ = – 1

$\frac{(\tan ^2 \theta - \sec ^2 \theta)} {(\tan \theta - \sec \theta)}$ = m

$\frac{- 1} {(\tan \theta - \sec \theta)}$ = m

∴ tan θ – sec θ = $\frac{- 1}{m}$ …… (ii)

By subtracting equation (ii) from equation (i), we get:

(tan θ + sec θ) – (tan θ – sec θ) = m – $(\frac{- 1}{m})$

∴ 2 sec θ = m + $\frac{1}{m} = \frac{m^2 + 1}{m}$

∴ sec θ = $\frac{m^2 + 1}{2m}$

Hence Proved !

Please do press “Heart” button if you liked the solution.

Contact us