The value of $$\frac{{{{\sec }^2}\theta }}{{{\text{cose}}{{\text{c}}^2}\theta }} + \frac{{{\text{cose}}{{\text{c}}^2}\theta }}{{{{\sec }^2}\theta }} - \left( {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } \right){\text{is:}}$$
A. 1
B. 0
C. -2
D. 2
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{{\sec }^2}\theta }}{{{\text{cose}}{{\text{c}}^2}\theta }} + \frac{{{\text{cose}}{{\text{c}}^2}\theta }}{{{{\sec }^2}\theta }} - \left( {{{\sec }^2}\theta + {\text{cose}}{{\text{c}}^2}\theta } \right) \cr & = \frac{{{{\sec }^4}\theta + {{\cos }^4}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }} - \left( {\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }}} \right) \cr & = \frac{{\left( {{{\sin }^4}\theta - {{\sin }^2}\theta } \right) + \left( {{{\cos }^4}\theta - {{\cos }^2}\theta } \right)}}{{{{\sin }^2}\theta .{{\cos }^2}\theta }} \cr & = \frac{{ - {{\sin }^2}\theta .{{\cos }^2}\theta - {{\sin }^2}\theta .{{\cos }^2}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }} \cr & = - 2 \cr & {\bf{Alternate:}} \cr & {\text{Put }}\theta = {45^ \circ } \cr & 1 + 1 - \left( {2 + 2} \right) = - 2 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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