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