If sec θ + tan θ = 1/3, then sec θ – tan θ = ………A) 3B) 1

1.	If sec θ + tan θ = 1/3, then sec θ – tan θ = ………A) 3B) 1/3C) 1D) 0
Answer» Correct option is: A) 3 sec \(\theta\) + tan \(\theta\) = \(\frac{1}{3}\) = ( sec \(\theta\) + tan \(\theta\)) ( sec \(\theta\) - tan \(\theta\)) = \(\frac{1}{3}\) ( sec \(\theta\) - tan \(\theta\)) (Multiplying both sides by (sec \(\theta\) - tan \(\theta\))) = \(sec^2\theta - tan^2\theta = \frac 13 ( sec\, \theta - tan \, \theta)\) (\(\because\) (a + b ) (a - b) = \(a^2-b^2\)) = sec \(\theta\) - tan \(\theta\) = 3 (\(sec^2 \theta - tan^2\theta\)) = 3 (\(\because\) \(Sec^2 \theta - tan ^2 \theta = 1\)) Hence, if sec \(\theta\) + tan \(\theta\) = \(\frac{1}{3}\) then sec \(\theta\) - tan \(\theta\) = 3 Correct option is: A) 3

If sec θ + tan θ = 1/3, then sec θ – tan θ = ………A) 3B) 1/3C) 1D) 0

Answer»

Correct option is: A) 3

sec \(\theta\) + tan \(\theta\) = \(\frac{1}{3}\)

= ( sec \(\theta\) + tan \(\theta\)) ( sec \(\theta\) - tan \(\theta\)) = \(\frac{1}{3}\) ( sec \(\theta\) - tan \(\theta\))

(Multiplying both sides by (sec \(\theta\) - tan \(\theta\)))

= \(sec^2\theta - tan^2\theta = \frac 13 ( sec\, \theta - tan \, \theta)\) (\(\because\) (a + b ) (a - b) = \(a^2-b^2\))

= sec \(\theta\) - tan \(\theta\) = 3 (\(sec^2 \theta - tan^2\theta\)) = 3 (\(\because\) \(Sec^2 \theta - tan ^2 \theta = 1\))

Hence, if sec \(\theta\) + tan \(\theta\) = \(\frac{1}{3}\) then sec \(\theta\) - tan \(\theta\) = 3

Correct option is: A) 3