If 3sec2θ + tanθ - 7 = 0, 0° < θ < 90°, then what is the value of $$\left( {\frac{{2\sin \theta + 3\cos \theta }}{{{\text{cosec}}\,\theta + \sec \theta }}} \right)?$$
A. 10
B. $$\frac{5}{2}$$
C. $$\frac{5}{4}$$
D. 4
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & 3{\sec ^2}\theta + \tan \theta - 7 = 0 \cr & \Rightarrow 3\left( {1 + {{\tan }^2}\theta } \right) + \tan \theta - 7 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + \tan \theta - 4 = 0 \cr & \Rightarrow 3{\tan ^2}\theta + 4\tan \theta - 3\tan \theta - 4 = 0 \cr & \Rightarrow \tan \theta \left( {3\tan \theta + 4} \right) - 1\left( {3\tan \theta + 4} \right) = 0 \cr & \Rightarrow \left( {3\tan \theta + 4} \right)\left( {\tan \theta - 1} \right) = 0 \cr & \Rightarrow \tan \theta = 1\,\,\,\,\,\,\,\,\,\therefore \theta = {45^ \circ } \cr & \therefore \,\frac{{2\sin \theta + 3\cos \theta }}{{{\text{cosec}}\,\theta + \sec \theta }} \cr & = \frac{{2\left( {\frac{1}{{\sqrt 2 }}} \right) + 3\left( {\frac{1}{{\sqrt 2 }}} \right)}}{{\sqrt 2 + \sqrt 2 }} \cr & = \frac{{\frac{5}{{\sqrt 2 }}}}{{2\sqrt 2 }} \cr & = \frac{5}{4} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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