What is the value of $$\frac{{\cos {{50}^ \circ }}}{{\sin {{40}^ \circ }}} + \frac{{3{\text{cosec}}\,{\text{8}}{0^ \circ }}}{{\sec {{10}^ \circ }}} - 2\cos {50^ \circ } \cdot {\text{cosec}}\,{40^ \circ }?$$
A. 2
B. 4
C. 5
D. 3
Answer: Option A
Solution(By Examveda Team)
$$\eqalign{ & \frac{{\cos {{50}^ \circ }}}{{\sin {{40}^ \circ }}} + \frac{{3{\text{cosec}}\,{\text{8}}{0^ \circ }}}{{\sec {{10}^ \circ }}} - 2\cos {50^ \circ } \cdot {\text{cosec}}\,{40^ \circ }{\text{ angle of sum}} = {90^ \circ } \cr & {\text{then,}} \cr & = \frac{{\sin {{40}^ \circ }}}{{\sin {{40}^ \circ }}} + \frac{{3{\text{cosec}}\,{\text{5}}{0^ \circ }}}{{{\text{cosec}}\,{\text{8}}{0^ \circ }}} - 2\sin {40^ \circ } \cdot {\text{cosec}}\,{10^ \circ } \cr & = 1 + 3 - 2 \cr & = 2 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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