Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A =

[[1]]

(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ) 

[[2]]

(iii) (sec A + tan A) (1 – sin A) =

[[3]]

(iv) $\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}$

[[4]]

(i) Since 9 is common to both terms, taking it out,

9 sec²A – 9 tan²A

= 9 (sec²A – tan²A)

As we know, $se{c}^{2}A – {\tan }^{2}A = 1$

9 (sec²A – tan²A) = 9×1 = 9             

Therefore, 9 sec²A – 9 tan²A = 9

So, option (B) is correct.

(ii) As we know that, $\tan \theta =\frac{ \sin \theta }{\cos \theta }$

$sec \theta =\frac{1}{\cos \theta }$

$cot \theta =\frac{\cos \theta }{\sin \theta }$

$\cos ec \theta = \frac{1}{\sin \theta }$

Substitute the obtained values into the given expression, we get

$(1 + \tan \theta + sec \theta ) (1 + cot \theta – \cos ec \theta ) = \left(1 + \frac{\sin \theta }{\cos \theta } +\frac{1}{\cos \theta }\right) \left(1 + \frac{\cos \theta }{\sin \theta } – \frac{1}{\sin \theta }\right)$

Simplify it further,

$\left(1 + \frac{\sin \theta }{\cos \theta } +\frac{1}{\cos \theta }\right) \left(1 + \frac{\cos \theta }{\sin \theta } – \frac{1}{\sin \theta }\right)=\left(\frac{\cos \theta +\sin \theta +1}{\cos \theta } \right) \left( \frac{\cos \theta + \sin \theta -1}{\sin \theta }\right)$

$=\left(\frac{{\left(\cos \theta +\sin \theta \right)}^2-1^2}{\sin \theta \cos \theta } \right)=\left(\frac{{\left(\cos \theta \right)}^2+{\left(\sin \theta \right)}^2+2\sin \theta \cos \theta -1^2}{\sin \theta \cos \theta } \right)$

Using, ${\sin }^{2}A+{\cos }^{2}A=1$

$=\frac{1+2\sin \theta \cos \theta -1}{\sin \theta \cos \theta }=\frac{2\sin \theta \cos \theta }{\sin \theta \cos \theta }=2$

So, option (C) is correct. 

(iii) As we know that, $\tan \theta =\frac{ \sin \theta }{\cos \theta }$

$sec \theta =\frac{1}{\cos \theta }$

Substitute the obtained values into the given expression, we get

$(secA + \tan A) (1 – \sin A)= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A} \right)\left(1 – \sin A\right)$

$= \left( \frac{1+\sin A}{\cos A} \right)\left(1 – \sin A\right)$

$= \left( \frac{1^2-{\left(\sin A\right)}^2}{\cos A} \right)$

Using, ${\sin }^{2}A+{\cos }^{2}A=1$

$\Rightarrow \left( \frac{1^2-{\left(\sin A\right)}^2}{\cos A} \right)=\left( \frac{{\left(\cos A\right)}^2}{\cos A} \right)=\cos A$

So, option (D) is correct.

(iv) As we know,  $\tan A=\frac{1}{cot A}$

Substitute the obtained values into the given expression, we get

$\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}=\frac{1+\left({\displaystyle \frac{1}{co{t}^{2}A}}\right)}{1+co{t}^{2}A}=\frac{{\displaystyle \frac{1+co{t}^{2}A}{co{t}^{2}A}}}{1+co{t}^{2}A}$

Simplify it further,

$\Rightarrow \frac{{\displaystyle \frac{1+co{t}^{2}A}{co{t}^{2}A}}}{1+co{t}^{2}A}=\frac{1}{co{t}^{2}A}=\tan { }^{2}A$

So, option (D) is correct.