Question

If 3cos θ = 2 then (2sec² θ + 2tan² θ – 7) = ?
(a) 0
(b) 1
(c) 3
(4) 4

Answer 1

$$\begin{gathered} \mathrm{Given}:3\cos \theta =2 \\ \Rightarrow \cos \theta =\frac{2}{3} \\ \mathrm{Since},\cos \theta =\frac{B}{H} \\ \Rightarrow B=2\mathrm{and}H=3 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow P^2+2^2=3^2 \\ \Rightarrow P^2=9-4 \\ \Rightarrow P^2=5 \\ \Rightarrow P=\sqrt{5} \\ \mathrm{Therefore}, \\ \sec \theta =\frac{H}{B}=\frac{3}{2} \\ \tan \theta =\frac{P}{B}=\frac{\sqrt{5}}{2} \\ \mathrm{Now}, \\ \left(2{\sec }^2\theta +2{\tan }^2\theta -7\right)=\left(2{\left(\frac{3}{2}\right)}^2+2{\left(\frac{\sqrt{5}}{2}\right)}^2-7\right) \\ =\left(2\left(\frac{9}{4}\right)+2\left(\frac{5}{4}\right)-7\right) \\ =\left(\frac{9}{2}+\frac{5}{2}-7\right) \\ =\left(\frac{9+5}{2}-7\right) \\ =\left(\frac{14}{2}-7\right) \\ =\left(7-7\right) \\ =0 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{a}\right). \end{gathered}$$