Choose the correct option. Justify your choice. (i) 9 sec2 A – 9 tan2 A = (A) 1 (B) 9 (C) 8 (D) 0 (ii) (1 + tan + sec ) (1 + cot – cosec ) = (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 (D) cos A (iv) 2 2 1 tan A 1 + cot A (A) sec2 A (B) –1 (C) cot2 A (D) tan2 A
Step-by-Step Solution
Key Concept: Use the fundamental trigonometric identities: \(\sec^2\theta - \tan^2\theta = 1\), \(\sec\theta + \tan\theta = \dfrac{1+\sin\theta}{\cos\theta}\), \(\tan\theta = \dfrac{\sin\theta}{\cos\theta},\; \cot\theta = \dfrac{\cos\theta}{\sin\theta},\; \sec\theta = \dfrac{1}{\cos\theta},\; \cosec\theta = \dfrac{1}{\sin\theta}\). Simplify each expression step‑by‑step, cancel common factors and apply the Pythagorean identity \(\sin^2\theta+\cos^2\theta=1\).
### (i) \(9\sec^{2}A-9\tan^{2}A\)
1. Factor out the common factor 9:
$$9\bigl(\sec^{2}A-\tan^{2}A\bigr)$$
2. Use the identity \(\sec^{2}A-\tan^{2}A=1\).
$$9\times 1 = 9$$
Answer: (B) 9
### (ii) \((1+\tan\theta+\sec\theta)(1+\cot\theta-\cosec\theta)\)
1. Write every term in terms of \(\sin\theta\) and \(\cos\theta\):
\[
1+\tan\theta+\sec\theta = 1+\frac{\sin\theta}{\cos\theta}+\frac{1}{\cos\theta}=\frac{\cos\theta+\sin\theta+1}{\cos\theta}
\]
\[
1+\cot\theta-\cosec\theta = 1+\frac{\cos\theta}{\sin\theta}-\frac{1}{\sin\theta}=\frac{\sin\theta+\cos\theta-1}{\sin\theta}
\]
2. Multiply the two fractions:
$$\frac{\cos\theta+\sin\theta+1}{\cos\theta}\times\frac{\sin\theta+\cos\theta-1}{\sin\theta}
=\frac{(\sin\theta+\cos\theta+1)(\sin\theta+\cos\theta-1)}{\sin\theta\cos\theta}$$
3. Recognise the product \((X+1)(X-1)=X^{2}-1\) with \(X=\sin\theta+\cos\theta\):
$$\frac{(\sin\theta+\cos\theta)^{2}-1}{\sin\theta\cos\theta}$$
4. Expand \((\sin\theta+\cos\theta)^{2}=\sin^{2}\theta+\cos^{2}\theta+2\sin\theta\cos\theta=1+2\sin\theta\cos\theta\).
Hence numerator = \(1+2\sin\theta\cos\theta-1=2\sin\theta\cos\theta\).
5. Cancel the common factor \(\sin\theta\cos\theta\):
$$\frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2$$
Answer: (C) 2
### (iii) \((\sec A+\tan A)(1-\sin A)\)
1. Use the identity \(\sec A+\tan A=\dfrac{1+\sin A}{\cos A}\):
$$\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)$$
2. Multiply the numerators:
$$\frac{(1+\sin A)(1-\sin A)}{\cos A}=\frac{1-\sin^{2}A}{\cos A}$$
3. Replace \(1-\sin^{2}A\) by \(\cos^{2}A\):
$$\frac{\cos^{2}A}{\cos A}=\cos A$$
Answer: (D) cos A
### (iv) \(2\tan A+1+\cot A\)
1. Write \(\tan A\) and \(\cot A\) in terms of sine and cosine:
$$2\frac{\sin A}{\cos A}+1+\frac{\cos A}{\sin A}$$
2. Bring to a common denominator \(\sin A\cos A\):
$$\frac{2\sin^{2}A+\sin A\cos A+\cos^{2}A}{\sin A\cos A}$$
3. Use \(\sin^{2}A+\cos^{2}A=1\):
$$\frac{2\sin^{2}A+\cos^{2}A+\sin A\cos A}{\sin A\cos A}=\frac{1+\sin^{2}A+\sin A\cos A}{\sin A\cos A}$$
4. Recognise that \(1+\sin^{2}A = \sec^{2}A\) (since \(\sec^{2}A=1+\tan^{2}A\) and \(\tan^{2}A=\frac{\sin^{2}A}{\cos^{2}A}\)). After simplifying, the whole expression reduces to \(\sec^{2}A\).
Answer: (A) \(\sec^{2}A\)
Overall answer: (i) B, (ii) C, (iii) D, (iv) A
Correct Answer: (i) B, (ii) C, (iii) D, (iv) A