Question

If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.

Answer 1

Given,

tan θ + sec θ = l   ...(i)

⇒ `((tan theta + sec  theta)(sec theta - tan theta))/((sec theta - tan theta))` = l   ...[Multiply by (sec θ – tan θ) on numerator and denominator L.H.S]

⇒ `((sec^2 theta - tan^2 theta))/((sec theta - tan theta))` = l

⇒ `1/(sec theta - tan theta)` = l   ...[∵ sec²θ – tan²θ = 1]

⇒ sec θ – tan θ = `1/l`  ...(ii)

On adding equations (i) and (ii), we get

2 sec θ = `l + 1/l`

⇒ sec θ = `(l^2 + 1)/(2l)`  

Hence proved.