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) $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$
(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 (D) cos A
(iv) $\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=$
(A) $\sec ^{2} \mathrm{~A}$ (B) $-1$ (C) $\cot ^{2} A$ (D) $\tan ^{2} \mathrm{~A}$
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) $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$
(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 (D) cos A
(iv) $\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=$
(A) $\sec ^{2} \mathrm{~A}$ (B) $-1$ (C) $\cot ^{2} A$ (D) $\tan ^{2} \mathrm{~A}$
Solution:
(i) Correct option is (B).
$9 \sec ^{2} A-9 \tan ^{2} A=9\left(\sec ^{2} A-\tan ^{2} A\right)$
= 9 × 1 = 9.
(ii) Correct option is (C).
$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$
$=\left\{1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right\} \times\left\{1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right\}$
$=\left\{\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right\} \times\left\{\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right\}$
$=\frac{\{(\cos \theta+\sin \theta)+1\} \times\{(\cos \theta+\sin \theta)-1\}}{\cos \theta \times \sin \theta}$
$=\frac{(\cos \theta+\sin \theta)^{2}-(1)^{2}}{\cos \theta \times \sin \theta}$
$\left\{\because(a+b)(a-b)=a^{2}-b^{2}\right\}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}{\cos \theta \times \sin \theta}$
$=\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}=2$
(iii) Correct option is (D).
(secA + tanA)(1 – sinA)
$=\sec A-\tan A+\tan A-\frac{\sin ^{2} A}{\cos A}$
$=\frac{1}{\cos A}-\frac{\sin ^{2} A}{\cos A}=\frac{1-\sin ^{2} A}{\cos A}$
$=\frac{\cos ^{2} \mathbf{A}}{\cos \mathbf{A}}=\cos \mathbf{A}$
(iv) Correct option is (D).
$\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=\frac{\sec ^{2} \mathbf{A}}{\operatorname{cosec}^{2} \mathbf{A}}=\tan ^{2} \mathrm{~A}$
(i) Correct option is (B).
$9 \sec ^{2} A-9 \tan ^{2} A=9\left(\sec ^{2} A-\tan ^{2} A\right)$
= 9 × 1 = 9.
(ii) Correct option is (C).
$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$
$=\left\{1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right\} \times\left\{1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right\}$
$=\left\{\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right\} \times\left\{\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right\}$
$=\frac{\{(\cos \theta+\sin \theta)+1\} \times\{(\cos \theta+\sin \theta)-1\}}{\cos \theta \times \sin \theta}$
$=\frac{(\cos \theta+\sin \theta)^{2}-(1)^{2}}{\cos \theta \times \sin \theta}$
$\left\{\because(a+b)(a-b)=a^{2}-b^{2}\right\}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}{\cos \theta \times \sin \theta}$
$=\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}=2$
(iii) Correct option is (D).
(secA + tanA)(1 – sinA)
$=\sec A-\tan A+\tan A-\frac{\sin ^{2} A}{\cos A}$
$=\frac{1}{\cos A}-\frac{\sin ^{2} A}{\cos A}=\frac{1-\sin ^{2} A}{\cos A}$
$=\frac{\cos ^{2} \mathbf{A}}{\cos \mathbf{A}}=\cos \mathbf{A}$
(iv) Correct option is (D).
$\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=\frac{\sec ^{2} \mathbf{A}}{\operatorname{cosec}^{2} \mathbf{A}}=\tan ^{2} \mathrm{~A}$