Question

Q1.cos7a + cos3a - cos 5a - cos a /
sin 7a - sin 3a - sin 5a + sin a = cot 2a ​

Answer 1

Step-by-step explanation:

$\begin{gathered}\cos7a+\cos3a-\cos5a-\cos a\\\;\\=(\cos7a-\cos a)+(\cos3a-\cos5a)\\\;\\=(2\sin\frac{8a}{2}.\sin(\frac{-6a}{2}))+(2\sin\frac{8a}{2}.\sin\frac{2a}{2})\;\;\;\;\;\;[\because \cos C-\cos D=2\sin(\frac{C+D}{2})\sin(\frac{D-C}{2})]\\\;\\\;=-2\sin4a.\sin3a\;+\;2\sin4a.\sin a\end{gathered}$

and

$\begin{gathered}\sin7a-\sin3a-\sin5a+\sin a\\\;\\=(\sin7a+\sin a)-(\sin5a+\sin3a)\\\;\\=(2\sin\frac{8a}{2}.\cos(\frac{6a}{2}))+(2\sin\frac{8a}{2}.\cos\frac{2a}{2})\;\;\;\;\;\;[\because \sin C+\sin D=2\sin(\frac{C+D}{2})\cos(\frac{C-D}{2})]\\\;\\\;=2\sin4a.\cos3a\;-\;2\sin4a.\cos a\end{gathered}$

Now Coming to question,

$\begin{gathered}\frac{\cos7a+\cos3a-\cos5a-\cos a}{\sin7a-\sin3a-\sin5a+\sin a}\\\;\\=\frac{-2\sin4a\sin3a+2\sin4a\sin a}{2\sin4a\cos3a-2\sin4a\cos a}\\\;\\=\frac{-\sin4a(\sin3a-\sin a)}{\sin4a(\cos3a-\cos a)}\\\;\\=-\frac{\sin3a-\sin a}{\cos3a-\cos a}\\\;\\=\frac{-2\cos2a\sin a}{2\sin2a\sin(-a)}\\\;\\=\frac{-\cos2a\sin a}{-\sin2a\sin a}\\\;\\=\frac{\cos2a}{\sin2a}\\\;\\=\cot2a\end{gathered}$

Note:-

$\begin{gathered}1.\;\sin C-\sin D=2\cos(\frac{C+D}{2})\sin(\frac{C-D}{2})\\\;\\2.\;\cos C-\cos D=2\sin(\frac{C+D}{2})\sin(\frac{D-C}{2})\end{gathered}$