Question

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

Answer 1

$\begin{array}{c}a\sec \theta =1-b\tan \theta \\ a^2{\sec }^2\theta =1+b^2{\tan }^2\theta -2b\tan \theta ------\left(1\right)\\ Also,a^2{\sec }^2\theta =5+b^2{\tan }^2\theta -------\left(2\right)\\ From\left(1\right)and\left(2\right)\\ 1-2b\tan \theta =5\Rightarrow b\tan \theta =-2\\ \therefore a\sec \theta =3\Rightarrow a=3\cos \theta \\ a^2=9{\cos }^2\theta -------\left(3\right)\\ b\tan \theta =-2\Rightarrow b^2\frac{{\sin }^2\theta }{{\cos }^2\theta }=4\\ {\cos }^2\theta =\frac{b^2}{b^2+4}-------\left(4\right)\\ from\left(3\right)and\left(4\right),weget\\ a^2=\frac{9b^2}{b^2+4}\Rightarrow \left(b^2+4\right)a^2=9b^2\end{array}$