If 3 cos $\theta$ = 2 then (2 sec2$\theta$ + 2 tan2$\theta$ – 7) = ?

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KollegeApply · Accepted Answer

0 Explanation: cos $\theta$ = $\frac{2}{3}$ $\Rightarrow$ (AB)/(AC) = $\frac{2}{3}$ Let AB = 2k and AC = 3k. Then, BC2 = (AC2 – AB2) = (9k2 – 4k2) = 5k2 $\therefore$ BC = sqrt(5)k `2(sec^2$\theta$ + tan^2$\theta$) - 7 = 2[((AC)/(AB))^2 + ((BC)/(AB))^2] - 7 = 2[((3k)/(2k))^2 + ((sqrt(5)k)/(2k))^2] - 7 = {2($\frac{3}{2}$)^2 + 2((sqrt(5))/2)^2 - 7} = (2 xx $\frac{9}{4}$ + 2 xx $\frac{5}{4}$ - 7)` = 0

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