Question

$a\sec \theta +b\tan \theta =1,a^2{\sec }^2\theta -b^2{\tan }^2\theta =5$
Then $a^2(b^2+4)=$

• A. $3b^2$
• B. $9b^2$
• C. $b^2$
• D. $4b^2$

Answer 1

Answer: (B)

$9b^2$

$a\sec \theta +b\tan \theta =1$ ............ $\left(1\right)$

Squaring both the sides we get

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

$a^2{\sec }^2\theta -b^2{\tan }^2\theta =5$ ....... $\left(3\right)$

Adding equations $\left(2\right)$ and $\left(3\right)$ we get,

$2a^2{\sec }^2\theta +2ab\sec \theta \tan \theta =6$

$a^2{\sec }^2\theta +ab\sec \theta \tan \theta =3$

$a\sec \theta (a\sec \theta +b\tan \theta )=3$

$a\sec \theta =3$ .......... (Using eqn $\left(1\right)$)

$⟹a=3\cos \theta$ .......... $(\because \cos \theta =\frac{1}{\sec \theta })$

Substitute $a$ in $\left(1\right)$ we get

$\left(3\cos \theta \right)\sec \theta +b\tan \theta =1$

$3+b\tan \theta =1$

$btan\theta =-2$

$b=-2\cot \theta$ ............. $(\because \frac{1}{\tan \theta }=\cot \theta )$

$\therefore a^2(b^2+4)=(3\cos \theta {)}^2[{(-2\cot \theta )}^2+4]$

$=9{cos}^2\theta (4{cot}^2\theta +4)$

$=9{\cos }^2\theta \cdot 4({\cot }^2\theta +1)$

$=9{\cos }^2\theta \cdot 4{\text{cosec}}^2\theta$

$=36\times \frac{{\cos }^2\theta }{{\sin }^2\theta }$

$=36{\cot }^2\theta =9(-2cot\theta {)}^2=9b^2$

Hence, $a^2(b^2+4)=9b^2$