The numerical value of 1 + $$\frac{1}{{{\text{co}}{{\text{t}}^2}{{63}^ \circ }}}$$ - $${\text{se}}{{\text{c}}^2}{27^ \circ }$$ + $$\frac{1}{{{{\sin }^2}{{63}^ \circ }}}$$ - $${\text{cose}}{{\text{c}}^2}{27^ \circ }$$ is?
A. 1
B. 2
C. -1
D. 0
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {\text{1 + }}\frac{1}{{{\text{co}}{{\text{t}}^2}{{63}^ \circ }}} - {\text{se}}{{\text{c}}^2}{27^ \circ }{\text{ + }}\frac{1}{{{{\sin }^2}{{63}^ \circ }}} - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{ta}}{{\text{n}}^2}{63^ \circ } - {\text{se}}{{\text{c}}^2}{27^ \circ } + {\text{cose}}{{\text{c}}^2}{63^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\sec ^2}{27^ \circ } + {\text{se}}{{\text{c}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 + {\text{co}}{{\text{t}}^2}{27^ \circ } - {\text{cose}}{{\text{c}}^2}{27^ \circ } \cr & \Rightarrow 1 - 1 \cr & \Rightarrow 0 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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