Question

$a\sec \theta +b\tan \theta =1,a\sec \theta -b\tan \theta =5\Rightarrow a^2(b^2+4)=?$

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

Answer 1

The correct option is B $9b^2$

$$\begin{array}{c}a\sec \theta +b\tan \theta =1,a\sec \theta -b\tan \theta =5\\ \frac{a+b\sin \theta }{\cos \theta }=1\\ a+b\sin \theta =\cos \theta .....\left(i\right)\\ \frac{a-b\sin \theta }{\cos \theta }=5\\ a-b\sin \theta =5\cos \theta .......\left(ii\right)\\ \\ addingequation\left(i\right)and\left(ii\right)\\ a+b\sin \theta =\cos \theta \\ a-b\sin \theta =5\cos \theta \\ 2a=6\cos \theta \\ a=3co\theta ,a^2=q{\cos }^2\theta \\ \\ puttingthevalueobtained\left(ii\right)\\ 3\cos \theta -b\sin \theta =5\cos \theta \\ -2\cos \theta =b\sin \theta \\ -2\cot \theta =b,b^2=4{\cot }^2\theta \\ then{a}^2\left(b^2+4\right)=9{\cos }^2\theta \left[4\left(1+{\cot }^2\theta \right)\right]\\ =36{\cos }^2\theta .\cos e{c}^2\theta \\ =36{\left(\frac{\cos \theta }{\sin \theta }\right)}^2\\ =36\times {\cot }^2\theta \left[\cdot .\cdot {\cot }^2\theta =\frac{b^2}{4}\right]\\ =\frac{36}{4}{b}^2\\ =9b^2\end{array}$$