(sin 7A-sin5A-sin3A+sinA)/(cos7A+cos3A-c os5A+cosA)=tan2A

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Step-by-step explanation:

\begin{gathered} \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ LHS = \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} \\ \\ = \frac{ \sin5A \:\cos A - \cos5A \: \sin A}{\sin A \:\cos A } \\ \\ = \frac{ \sin(5A -A )}{\sin A \:\cos A} \\ \\ = \frac{ \sin 4A}{\sin A \:\cos A} \\ \\ = \frac{ 2\sin 2A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 2 \times 2\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ = \frac{ 4\sin A \: \cos A \: \cos 2A}{\sin A \:\cos A} \\ \\ =4\cos 2A \\ \\ = RHS \\ \\ thus \\ \\ \frac{ \sin5A}{\sin A} - \frac{ \cos5A}{\cos A} = 4\cos2A \\ \\ hence \: proved\end{gathered}

sinA

sin5A

−

cosA

cos5A

=4cos2A

LHS=

sinA

sin5A

−

cosA

cos5A

=

sinAcosA

sin5AcosA−cos5AsinA

=

sinAcosA

sin(5A−A)

=

sinAcosA

sin4A

=

sinAcosA

2sin2Acos2A

=

sinAcosA

2×2sinAcosAcos2A

=

sinAcosA

4sinAcosAcos2A

=4cos2A

=RHS

thus

sinA

sin5A

−

cosA

cos5A

=4cos2A

henceproved

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