Question

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

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

Answer 1

Answer: (A)

$9b^2$

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

$\Rightarrow a^2{\sec }^2\theta =1+b^2{\tan }^2\theta -2b\tan \theta$....(i)

Also, $a^2{\sec }^2\theta =5+b^2{\tan }^2\theta$....(ii)

From (i) & (ii)

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

$\therefore a\sec \theta =3\Rightarrow a=3\cos \theta$

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

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

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

From (iii) and (iv), we get

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