Choose the correct option. Justify your choice :
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) $\tan ^{2} \mathrm{~A}$


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).

$\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=\frac{\sec ^{2} \mathbf{A}}{\operatorname{cosec}^{2} \mathbf{A}}=\tan ^{2} \mathrm{~A}$
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