?
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?
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& = {\sec ^2}{17^ \circ } - \frac{1}{{{{\tan }^2}{{73}^ \circ }}} - \sin {17^ \circ }\sec {73^ \circ } \cr
& = {\sec ^2}{17^ \circ } - {\cot ^2}{73^ \circ } - \sin {17^ \circ }\sec \left( {{{90}^ \circ } - {{17}^ \circ }} \right) \cr
& = {\sec ^2}{17^ \circ } - {\cot ^2}\left( {{{90}^ \circ } - {{17}^ \circ }} \right) - \sin {17^ \circ }\cos ec{17^ \circ } \cr
& = {\sec ^2}{17^ \circ } - {\tan ^2}{17^ \circ } - 1 \cr
& = 1 - 1\left[ {\because {{\sec }^2}\theta - {{\tan }^2}\theta = 1} \right] \cr
& = 0 \cr} $$
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