Question

If $a\sec \theta +b\tan \theta =1$ and $a^2{\sec }^2\theta -b^2{\tan }^2\theta$ $=5$, than $a^2{b}^2+4a^2=$

• A. $9b^2$
• B. $\frac{9}{a^2}$
• C. $\frac{-2}{b}$
• D. $9$

Answer 1

The correct option is A $9b^2$

Given:

$a\sec \theta +b\tan \theta =\mathrm{1.........}\left(i\right)$

and $a^2{\sec }^2\theta -b^2{\tan }^2\theta =\mathrm{5.............}\left(ii\right)$

From (i)

$a\sec \theta =1-b\tan \theta$

squaring both sides,

$a^2{\sec }^2\theta =1+b^2{\tan }^2\theta -2b\tan \theta ..........\left(iii\right)$

From (ii)

$a^2{\sec }^2\theta =5+b^2{\tan }^2\theta .............\left(iv\right)$

$\therefore$ from (iii) and (iv)

$\Rightarrow 1-2b\tan \theta =5$

$\Rightarrow -2b\tan \theta =4$

$\Rightarrow b\tan \theta =-\mathrm{2..........}\left(v\right)$

$\therefore$ from (i)

$a\sec \theta -2=1$

$\Rightarrow a\sec \theta =3$

$\Rightarrow a=3\cos \theta$

squaring both sides,

$\Rightarrow a^2=9{\cos }^2\theta ..........\left(vi\right)$

now, from (v)

$b\tan \theta =-2$

squaring both sides,

$\Rightarrow b^2\frac{{\sin }^2\theta }{{\cos }^2\theta }=4$

$\Rightarrow 4{\cos }^2\theta =b^2{\sin }^2\theta$

$\Rightarrow 4{\cos }^2\theta =b^2(1-{\cos }^2\theta )$

$\Rightarrow 4{\cos }^2\theta +b^2{\cos }^2\theta =b^2$

$\Rightarrow {\cos }^2\theta (4+b^2)=b^2$

$\Rightarrow {\cos }^2\theta =\frac{b^2}{b^2+4}........\left(vii\right)$

From (vi) and (vii), we get

$a^2=\frac{9b^2}{b^2+4}$

$\Rightarrow (b^2+4)a^2=9b^2$

$\Rightarrow a^2{b}^2+4a^2=9b^2$

Hence, the required answer is $9b^2$