Choose the correct option. Justify your choice:
(i) B
(ii) C
(iii) D
(iv) D
We will use standard trigonometric identities to simplify each expression and then match it with the correct option.
Useful identities (keep in mind):
\(\sec^2 A - \tan^2 A = 1\), \(1+\tan^2 A = \sec^2 A\), \(1+\cot^2 A = \text{cosec }^2 A\),
\(\sec A + \tan A = \dfrac{1+\sin A}{\cos A}\), \(\text{cosec }\theta - \cot\theta = \dfrac{1-\cos\theta}{\sin\theta}\).
(i) \(9\sec^2 A - 9\tan^2 A\)
Take 9 common:
\(9\sec^2 A - 9\tan^2 A = 9(\sec^2 A - \tan^2 A)\).
Using \(\sec^2 A - \tan^2 A = 1\),
\(= 9 \times 1 = 9\).
Correct option: (B) 9.
(ii) \((1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta)\)
Step 1: Rewrite each bracket in a helpful way.
\(1+\tan\theta+\sec\theta = 1 + \dfrac{\sin\theta}{\cos\theta} + \dfrac{1}{\cos\theta} = 1 + \dfrac{\sin\theta+1}{\cos\theta}\)
\(= \dfrac{\cos\theta + \sin\theta + 1}{\cos\theta}\).
Also,
\(1+\cot\theta-\text{cosec }\theta = 1 + \dfrac{\cos\theta}{\sin\theta} - \dfrac{1}{\sin\theta} = 1 + \dfrac{\cos\theta-1}{\sin\theta}\)
\(= \dfrac{\sin\theta + \cos\theta - 1}{\sin\theta}\).
Step 2: Multiply them:
\((1+\tan\theta+\sec\theta)(1+\cot\theta-\text{cosec }\theta) = \dfrac{(\sin\theta+\cos\theta+1)(\sin\theta+\cos\theta-1)}{\sin\theta\cos\theta}\)
Step 3: Use \((x+1)(x-1)=x^2-1\) with \(x=\sin\theta+\cos\theta\):
\(= \dfrac{(\sin\theta+\cos\theta)^2 - 1}{\sin\theta\cos\theta}\)
Expand \((\sin\theta+\cos\theta)^2\):
\(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 + 2\sin\theta\cos\theta\).
So numerator becomes:
\((1 + 2\sin\theta\cos\theta) - 1 = 2\sin\theta\cos\theta\).
Therefore,
\(= \dfrac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2\).
Correct option: (C) 2.
(iii) \((\sec A + \tan A)(1-\sin A)\)
Step 1: Write \(\sec A + \tan A\) in terms of \(\sin A\) and \(\cos A\):
\(\sec A + \tan A = \dfrac{1}{\cos A} + \dfrac{\sin A}{\cos A} = \dfrac{1+\sin A}{\cos A}\).
Step 2: Multiply:
\((\sec A + \tan A)(1-\sin A) = \dfrac{1+\sin A}{\cos A}(1-\sin A)\)
\(= \dfrac{(1+\sin A)(1-\sin A)}{\cos A}\)
\(= \dfrac{1-\sin^2 A}{\cos A}\).
Step 3: Use \(1-\sin^2 A = \cos^2 A\):
\(= \dfrac{\cos^2 A}{\cos A} = \cos A\).
Correct option: (D) \(\cos A\).
(iv) \(\dfrac{1+\tan^2 A}{1+\cot^2 A}\)
Use identities:
\(1+\tan^2 A = \sec^2 A\) and \(1+\cot^2 A = \text{cosec }^2 A\).
So,
\(\dfrac{1+\tan^2 A}{1+\cot^2 A} = \dfrac{\sec^2 A}{\text{cosec }^2 A}\).
Now write in \(\sin\) and \(\cos\):
\(\dfrac{\sec^2 A}{\text{cosec }^2 A} = \dfrac{\frac{1}{\cos^2 A}}{\frac{1}{\sin^2 A}} = \dfrac{\sin^2 A}{\cos^2 A} = \tan^2 A\).
Correct option: (D) \(\tan^2 A\).
Final Answers: (i) B, (ii) C, (iii) D, (iv) D.