If 3(cot2θ - cos2θ) = cos2θ, 0°< θ < 90°, then the value of (tan2θ + cosec2θ + sin2θ) is:
A. $$\frac{{13}}{3}$$
B. $$\frac{{61}}{{12}}$$
C. $$\frac{{25}}{{12}}$$
D. $$\frac{{15}}{4}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & 3\left( {{{\cot }^2}\theta - {{\cos }^2}\theta } \right) = {\cos ^2}\theta \cr & 3{\cot ^2}\theta = {\cos ^2}\theta + 3{\cos ^2}\theta \cr & 3{\cot ^2}\theta = 4{\cos ^2}\theta \cr & \frac{3}{4} = {\sin ^2}\theta \cr & \sin \theta = \sin {60^ \circ } \cr & \theta = {60^ \circ } \cr & {\tan ^2}\theta + {\text{cose}}{{\text{c}}^2}\theta + {\sin ^2}\theta \cr & = {\tan ^2}{60^ \circ } + {\text{cose}}{{\text{c}}^2}{60^ \circ } + {\sin ^2}{60^ \circ } \cr & = 3 + \frac{4}{3} + \frac{3}{4} \cr & = \frac{{36 + 16 + 9}}{{12}} \cr & = \frac{{61}}{{12}} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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