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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) $(1 + tan\ θ + sec\ θ) (1 + cot\ θ - \operatorname{cosec} θ ) =$
(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) $\frac{1+tan^2\ A}{1+cot^2\ A}$
(A) $sec^2\ A$ (B) $-1$ (C) $cot^2\ A$ (D) $tan^2\ A$.
To do:
We have to choose the correct option and justify it.
Solution:
(i) We know that,
$sec^2\ \theta - tan^2\ \theta =1$
$9\ sec^2\ A - 9\ tan^2 A = 9(sec^2\ A - tan^2 A)$
$=9(1)$
$=9$
The correct option is B.
(ii)$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=(\frac{1}{1}+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta})(\frac{1}{1}+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta})$
$=(\frac{\cos \theta+\sin \theta+1}{\cos \theta})(\frac{\sin \theta+\cos \theta-1}{\sin \theta})$
$=\frac{(\cos \theta+\sin \theta)^{2}-(1)^{2}}{\cos \theta \sin \theta}$
$=\frac{\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}$
$=\frac{1+2 \cos \theta \sin \theta-1}{\cos \theta \sin \theta}$
$=\frac{2 \cos \theta \sin \theta}{\cos \theta \sin \theta}$
$=2$
The correct option is $C$.
(iii) $(\sec \mathrm{A}+\tan \mathrm{A})(1-\sin \mathrm{A})=(\frac{1}{\cos \mathrm{A}}+\frac{\sin \mathrm{A}}{\cos \mathrm{A}})(\frac{1-\sin \mathrm{A}}{1})$
$=(\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}})(\frac{1-\sin \mathrm{A}}{1})$
$=\frac{(1)^{2}-(\sin \mathrm{A})^{2}}{\cos \mathrm{A}}$
$=\frac{1-\sin ^{2} \mathrm{~A}}{\cos \mathrm{A}}$
$=\frac{\cos ^{2} \mathrm{~A}}{\cos \mathrm{A}}$
$=\cos \mathrm{A}$
The correct option is D.
(iv) $\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}=\frac{\sec ^{2} \mathrm{~A}}{\operatorname{cosec}^{2} \mathrm{~A}}$
$=\frac{\frac{1}{\cos ^{2} \mathrm{~A}}}{\frac{1}{\sin ^{2} \mathrm{~A}}}$
$=\frac{1}{\cos ^{2} \mathrm{~A}} \times \frac{\sin ^{2} \mathrm{~A}}{1}$
$=\frac{\sin ^{2} \mathrm{~A}}{\cos ^{2} \mathrm{~A}}$
$=\tan ^{2} \mathrm{~A}$
The correct option is D.
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