Question :
Choose the correct option. Justify your choice :
(i) \(9 sec^2 A – 9 tan^2 A\) =
(a) 1
(b) 9
(c) 8
(d) 0
(ii) \((1 – tan \theta + sec \theta)\)\((1 + cot \theta – cosec \theta)\) =
(a) 0
(b) 1
(c) 2
(d) -1
(iii) (sec A + tan A)(1 – sin A) =
(a) sec A
(b) sin A
(c) cosec A
(c) cos A
(iv) \(1 + tan^2 A\over 1 + cot^2 A\) =
(a) \(sec^2 A\)
(b) -1
(c) \(cot^2 A\)
(d) \(tan^2 A\)
Solution :
(i) (b) Because : \(9 sec^2 A – 9 tan^2 A\) = 9(\(9 sec^2 A – 9 tan^2 A\)) = 9(1) = 9
(ii) (c) Because : \((1 – tan \theta\) + sec \theta)\)\((1 + cot \theta – cosec \theta)\)
= (1 + \(sin\theta\over cos\theta\) + \(1\over cos\theta\))(1 + \(cos\theta\over sin\theta\) – \(1\over sin\theta\))
= \(({cos\theta +sin\theta})^2 – 1\over cos\theta sin\theta \)
= \(1 + 2cos\theta sin\theta – 1\over sin\theta cos\theta\) = \(2sin\theta cos\theta\over sin\theta cos\theta\) = 2
(iii) (d) Because
(sec A + tan A)(1 – sin A) = (\(1\over cos A\) + \(sin A\over cos A\))(1 – sin A) = (\(1 + sin A\over cos A\))(1 – sin A)
= \(1 – sin^2A\over cos A\) = \(cos^2 A\over cos A\) = cos A
(iv) (d) Because
\(1 + tan^2 A\over 1 + cot^2 A\) = \(1 + tan^2 A\over 1 + {1\over tan^2 A}\) = (\(1 + tan^2 A\)) \(\times\) \(tan^2 A\over 1 + tan^2 A\) = \(tan^2 A\)
