Sec θ (1 – Sin θ) (Sec θ + Tan θ) = A) 0 B) 1 C) 2 D) -1

1.	Sec θ (1 – Sin θ) (Sec θ + Tan θ) = A) 0 B) 1 C) 2 D) -1
Answer» Correct option is: B) 1 Sec \(\theta\) (1 – Sin \(\theta\)) (Sec \(\theta\) + Tan \(\theta\)) = sec \(\theta\) (1- sin \(\theta\)) (\(\frac 1{cos\, \theta} + \frac {sin \, \theta}{cos \, \theta}\)) = sec \(\theta\) ( 1- sin \(\theta\)) . \(\frac {1+sin\, \theta}{cos \, \theta}\) = \(\frac {1-sin^2\theta}{cos^2\theta} \) (\(\because\) sec \(\theta\) = \(\frac 1{cos\, \theta}\) ) = \(\frac {cos^2\theta}{cos^2\theta} = 1 \) (\(\because\) \(1- sin^2\theta = cos^2\theta\)) Correct option is: B) 1

Answer»

Correct option is: B) 1

Sec \(\theta\) (1 – Sin \(\theta\)) (Sec \(\theta\) + Tan \(\theta\))

= sec \(\theta\) (1- sin \(\theta\)) (\(\frac 1{cos\, \theta} + \frac {sin \, \theta}{cos \, \theta}\))

= sec \(\theta\) ( 1- sin \(\theta\)) . \(\frac {1+sin\, \theta}{cos \, \theta}\)

= \(\frac {1-sin^2\theta}{cos^2\theta} \) (\(\because\) sec \(\theta\) = \(\frac 1{cos\, \theta}\) )

= \(\frac {cos^2\theta}{cos^2\theta} = 1 \) (\(\because\) \(1- sin^2\theta = cos^2\theta\))

Correct option is: B) 1

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