The value of $${\text{cot1}}{{\text{7}}^ \circ }$$ $$\left( {\cot {{73}^ \circ }\,{{\cos }^2}{{22}^ \circ }\, + \frac{1}{{\cot {{17}^ \circ }se{c^2}{{68}^ \circ }}}} \right)$$ is?
A. 0
B. 1
C. 27
D. $$\sqrt 3 $$
Answer: Option B
Solution(By Examveda Team)
$$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {\cot {{73}^ \circ }{{\cos }^2}{{22}^ \circ } + \frac{1}{{\cot {{17}^ \circ }se{c^2}{{68}^ \circ }}}} \right)$$$$ \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left[ {\cot \left( {{{90}^ \circ } - {{17}^ \circ }} \right){{\cos }^2}\left( {{{90}^ \circ } - {{68}^ \circ }} \right) + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right]$$
$$\eqalign{ & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\left( {tan{{17}^ \circ }si{n^2}{{68}^ \circ } + \tan {{17}^ \circ }{\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow {\text{cot1}}{{\text{7}}^ \circ }\tan {17^ \circ }\left( {si{n^2}{{68}^ \circ } + {\text{co}}{{\text{s}}^2}{{68}^ \circ }} \right) \cr & \Rightarrow 1\left( 1 \right) \cr & \Rightarrow 1 \cr} $$
Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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