Question

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

Answer 1

Step-by-step explanation:

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

and

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

Now Coming to question,

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

Note:-

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

Answer 2

$\huge \bold \color{green}{ \underline{ \underline \red{required \: answer : }}}$

$\bold{ \underline{ \underline{LHS }}} = \frac{ \cos7A + \cos3A - \cos5A - \cos \: A}{ \sin7A - \sin3A \: - \sin5A + \sin \: A}$

$\bold{ \underline{ \underline{RHS}}} = \cot2A$

$\huge\bold{{ \underline { \underline{solution}}}}$

first we consider LHS

$\bold{ = \frac{ \cos7A + \cos3A - \cos5A - \cos \: A}{ \sin7A - \sin3A \: - \sin5A + \sin \: A}}$

$\bold{= \frac{ - 2 \sin4A\sin3A + 2 \sin4A \sin \: A}{2 \sin4A \cos3A - 2 \sin4A \cos \: A}}$

$= \frac{ - \sin4A( \sin3A - \sin \: A)}{ \sin4A( \cos3A - \cos \: A)}$

$= \frac{ - \sin3A - \sin \: A}{ \cos3A - \cos \: A}$

$= \frac{ - 2 \cos2A \sin \: A}{2 \sin2A \sin\:( - A)}$

$= \frac{ - 2 \cos2A \sin \: A}{ - 2 \sin2A \sin \: A}$

$= \frac{ \cos2A}{ \sin2A}$

$= \cot2A$

$\huge \bold{so \: LHS = RHS}$

$\bold \color{blue}{hence \: proved}$