Task. Prove
cot ( a − b ) − tan ( a − b ) = 2 sin 2 a sin 2 a + sin 2 b \cot (a - b) - \tan (a - b) = \frac {2 \sin 2 a}{\sin 2 a + \sin 2 b} cot ( a − b ) − tan ( a − b ) = sin 2 a + sin 2 b 2 sin 2 a
Proof. This identity is not correct. For instance let a = π 3 a = \frac{\pi}{3} a = 3 π b = π 6 b = \frac{\pi}{6} b = 6 π
a − b = π 6 , a - b = \frac {\pi}{6}, a − b = 6 π , cot ( a − b ) = 3 , tan ( a − b ) = 1 3 , \cot (a - b) = \sqrt {3}, \qquad \tan (a - b) = \frac {1}{\sqrt {3}}, cot ( a − b ) = 3 , tan ( a − b ) = 3 1 , sin 2 a = sin 2 π 3 = 3 2 , sin 2 b = sin π 3 = 3 2 , \sin 2 a = \sin \frac {2 \pi}{3} = \frac {\sqrt {3}}{2}, \qquad \sin 2 b = \sin \frac {\pi}{3} = \frac {\sqrt {3}}{2}, sin 2 a = sin 3 2 π = 2 3 , sin 2 b = sin 3 π = 2 3 ,
so
cot ( a − b ) − tan ( a − b ) = 3 − 1 3 = 3 − 1 3 = 2 3 , \cot (a - b) - \tan (a - b) = \sqrt {3} - \frac {1}{\sqrt {3}} = \frac {3 - 1}{\sqrt {3}} = \frac {2}{\sqrt {3}}, cot ( a − b ) − tan ( a − b ) = 3 − 3 1 = 3 3 − 1 = 3 2 ,
while
2 sin 2 a sin 2 a + sin 2 b = 2 ⋅ 3 2 3 2 + 3 2 = 3 3 = 1 ≠ 2 3 . \frac {2 \sin 2 a}{\sin 2 a + \sin 2 b} = \frac {2 \cdot \frac {\sqrt {3}}{2}}{\frac {\sqrt {3}}{2} + \frac {\sqrt {3}}{2}} = \frac {\sqrt {3}}{\sqrt {3}} = 1 \neq \frac {2}{\sqrt {3}}. sin 2 a + sin 2 b 2 sin 2 a = 2 3 + 2 3 2 ⋅ 2 3 = 3 3 = 1 = 3 2 .
Notice that the left hand side can be simplified as follows:
cot ( a − b ) − tan ( a − b ) = cos ( a − b ) sin ( a − b ) − sin ( a − b ) cos ( a − b ) = cos 2 ( a − b ) − sin 2 ( a − b ) sin ( a − b ) cos ( a − b ) = = 2 cos 2 ( a − b ) 2 sin ( a − b ) cos ( a − b ) = 2 cos 2 ( a − b ) 2 sin ( a − b ) cos ( a − b ) = = 2 cos 2 ( a − b ) sin 2 ( a − b ) = 2 cot 2 ( a − b ) . \begin{array}{l}
\cot (a - b) - \tan (a - b) = \frac {\cos (a - b)}{\sin (a - b)} - \frac {\sin (a - b)}{\cos (a - b)} = \frac {\cos^ {2} (a - b) - \sin^ {2} (a - b)}{\sin (a - b) \cos (a - b)} = \\
= \frac {2 \cos 2 (a - b)}{2 \sin (a - b) \cos (a - b)} = \frac {2 \cos 2 (a - b)}{2 \sin (a - b) \cos (a - b)} = \\
= \frac {2 \cos 2 (a - b)}{\sin 2 (a - b)} = 2 \cot 2 (a - b).
\end{array} cot ( a − b ) − tan ( a − b ) = s i n ( a − b ) c o s ( a − b ) − c o s ( a − b ) s i n ( a − b ) = s i n ( a − b ) c o s ( a − b ) c o s 2 ( a − b ) − s i n 2 ( a − b ) = = 2 s i n ( a − b ) c o s ( a − b ) 2 c o s 2 ( a − b ) = 2 s i n ( a − b ) c o s ( a − b ) 2 c o s 2 ( a − b ) = = s i n 2 ( a − b ) 2 c o s 2 ( a − b ) = 2 cot 2 ( a − b ) . 
#340153 on Dec 2023