If $$2\frac{{{{\cos }^2}x - {{\sec }^2}x}}{{{{\tan }^2}x}} = a + b\cos 2x,$$ then a, b = ?
A. $$\frac{{ - 3}}{2},\,\frac{{ - 1}}{2}$$
B. $$\frac{3}{2},\,\frac{1}{2}$$
C. -3, -1
D. 3, 1
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & 2\frac{{{{\cos }^2}x - {{\sec }^2}x}}{{{{\tan }^2}x}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{{{\cos }^2}x - \frac{1}{{{{\cos }^2}x}}}}{{\frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{{{\cos }^4}x - 1}}{{{{\cos }^2}x}} \times \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = a + b\cos 2x \cr & \Rightarrow 2\frac{{\left( {{{\cos }^2}x - 1} \right)\left( {{{\cos }^2}x + 1} \right)}}{{\left( {1 - {{\cos }^2}x} \right)}} = a + b\cos 2x \cr & \Rightarrow \frac{{ - 2\left( {1 - {{\cos }^2}x} \right)\left( {{{\cos }^2}x + 1} \right)}}{{\left( {1 - {{\cos }^2}x} \right)}} = a + b\cos 2x \cr & \Rightarrow - 2{\cos ^2}x - 2 = a + b\cos 2x \cr & \Rightarrow - 2 + 1 - 1 - 2{\cos ^2}x = a + b\cos 2x \cr & \Rightarrow - 3 - \left( {2{{\cos }^2}x - 1} \right) = a + b\cos 2x \cr & \Rightarrow - 3 - \cos 2x = a + b{\cos ^2}x \cr & a = - 3,\,\,b = - 1 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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