If A, B and C be the angles of a triangle, the incorrect relation is ?
A. $${\text{sin }}\left( {\frac{{{\text{A + B}}}}{2}} \right) = {\text{cos}}\frac{{\text{C}}}{2}$$
B. $${\text{cos }}\left( {\frac{{{\text{A + B}}}}{2}} \right) = {\text{sin}}\frac{{\text{C}}}{2}$$
C. $${\text{tan }}\left( {\frac{{{\text{A + B}}}}{2}} \right) = \sec \frac{{\text{C}}}{2}$$
D. $${\text{cot }}\left( {\frac{{{\text{A + B}}}}{2}} \right) = \tan \frac{{\text{C}}}{2}$$
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & {\text{A}} + {\text{B}} + {\text{C}} = \pi = {\text{18}}{0^ \circ }{\text{ }} \cr & \Rightarrow \frac{{{\text{A + B}}}}{2} = \frac{{{{180}^ \circ }}}{2} - \frac{{\text{C}}}{2} \cr & \Rightarrow {\text{sin }}\left( {\frac{{{\text{A + B}}}}{2}} \right) \cr & \Rightarrow {\text{sin }}\left( {\frac{\pi }{2} - \frac{{\text{C}}}{2}} \right) \cr & \Rightarrow {\text{cos}}\frac{{\text{C}}}{2} \cr & \cr & {\bf{Similarly:}} \cr & {\text{cos }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = {\text{sin}}\frac{{\text{C}}}{2} \cr & {\text{cot }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \tan \frac{{\text{C}}}{2} \cr & {\text{tan }}\left( {\frac{{{\text{A}} + {\text{B}}}}{2}} \right) = \cot \frac{{\text{C}}}{2} \cr & {\text{So, option C is incorrect}}{\text{.}} \cr} $$Related Questions on Trigonometry
A. x = -y
B. x > y
C. x = y
D. x < y
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