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) $(1+\tan \theta +\sec \theta )(1+\cot \theta -\text{cosec}\theta )$

(A) $0$

(B) $1$

(C) $2$

(D) $-1$

(iii) $(\sec A+\tan A)(1-\sin A)=$

(A) $\sec A$

(B) $\sin A$

(C) $\text{cosec}A$

(D) $\cos A$

(iv) $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=$

(A) ${\sec }^{2}A$

(B) $-1$

(C) ${\cot }^{2}A$

(D) ${\tan }^{2}A$

Answer 1

(i) Given $9{\sec }^{2}A-9{\tan }^{2}A$

$=9({\sec }^{2}A-{\tan }^{2}A)$
$=9\left(1\right)$ [Since, ${\sec }^{2}A-{\tan }^{2}A=1$]
$=9$
Hence, alternative (B) is correct.

(ii) Given $(1+\tan \theta +\sec \theta )(1+\cot \theta -\text{cosec}\theta )$
Since $\tan \theta =\frac{\sin \theta }{\cos \theta },\cot \theta =\frac{\cos \theta }{\sin \theta },\sec \theta =\frac{1}{\cos \theta }$ and $\text{cosec}\theta =\frac{1}{\sin \theta }$

So, given expression can be written as
$(1+\tan \theta +\sec \theta )(1+\cot \theta -\text{cosec}\theta )=(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta })+(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta })$
$=\left(\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right)+\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)$
$=\frac{(\cos \theta +\sin \theta {)}^2-(1{)}^2}{\sin \theta \cos \theta }$
$=\frac{{\sin }^2\theta +{\cos }^2\theta +2\cos \theta \sin \theta -1}{\sin \theta \cos \theta }$
$=\frac{1+2\cos \theta \sin \theta -1}{\sin \theta \cos \theta }$ [Since, ${\sin }^2\theta +{\cos }^2\theta =1$]
$=\frac{2\cos \theta \sin \theta }{\cos \theta \sin \theta }$
$=2$
Hence, alternative (C) is correct.

(iii) Given $(\sec A+\tan A)(1-\sin A)=(\frac{1}{\cos A}+\frac{\sin A}{\cos A})(1-\sin A)$
$=\frac{1+\sin A}{\cos A}(1-\sin A)$
$=\frac{1-{\sin }^{2}A}{\cos A}$
$=\frac{{\cos }^{2}A}{\cos A}$
$=\cos A$
Hence, alternative (D) is correct.

(iv) Given $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=\frac{1+(\frac{\sin A}{\cos A}{)}^2}{1+(\frac{\cos A}{\sin A}{)}^2}=$
$=\frac{\frac{{\cos }^{2}A+{\sin }^{2}A}{{\cos }^{2}A}}{\frac{{\sin }^{2}A+{\cos }^{2}A}{{\sin }^{2}A}}$
$=\frac{\frac{1}{{\cos }^{2}A}}{\frac{1}{{\sin }^{2}A}}$
$=\frac{{\sin }^{2}A}{{\cos }^{2}A}$
$={\tan }^{2}A$
Hence, alternative (D) is correct.