The value of $${\text{se}}{{\text{c}}^2}{17^ \circ }$$ - $$\frac{1}{{{\text{ta}}{{\text{n}}^2}{{73}^ \circ }}}$$ - $$\sin {17^ \circ }$$ $$\sec {73^ \circ }$$ is?
A. 1
B. 0
C. -1
D. 2
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr & = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr & = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr & = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr & = 0 \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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