Math, asked by Brainlybarbiedoll, 5 months ago

(v)
cos 7a+cos3a-cos 5a- cosa/
sin 7a -sin 3a -sin 5a + sin a
= cot 2a​

Answers

Answered by silu12
3

Answer:

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Answered by vanshikavikal448
26

\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}

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