A) \[\frac{361}{2397}\]
B) \[\frac{271}{979}\]
C) \[\frac{541}{979}\]
D) \[\frac{127}{979}\]
Correct Answer: B
Solution :
As, \[3\cos x=5\sin x\]\[\Rightarrow \]\[\frac{\sin x}{\cos x}=\frac{3}{5}\]\[\Rightarrow \]\[\tan x=\frac{3}{5}\] \[\frac{5\sin x-2{{\sec }^{3}}x+2\cos x}{2\sin x+2{{\sec }^{3}}x-2\cos x}\] On dividing numerator and denominator by cos x, we get \[\frac{5\tan x-2{{\sec }^{4}}x+2}{5\tan x+2{{\sec }^{3}}x-2}\] \[=\frac{5\tan x-2\,\,{{(1+{{\tan }^{2}}x)}^{2}}+2}{5\tan x+2\,\,{{(1+{{\tan }^{2}}x)}^{2}}-2}\] \[=\frac{5\times \frac{3}{5}-2{{\left( 1+\frac{9}{25} \right)}^{2}}+2}{5\times \frac{3}{5}+2{{\left( 1+\frac{9}{25} \right)}^{2}}-2}=\frac{3+2-2\times \frac{1156}{625}}{3-2+2\times \frac{1156}{625}}\] \[=\frac{3125-2312}{625+2312}=\frac{813}{2937}=\frac{271}{979}\]
You need to login to perform this action.
You will be redirected in
3 sec
