Prove that $\frac{sin4A + sin5A + sin6A}{cos4A + cos5A + cos6A} = tan5A$

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Prove that

$\frac{sin4A + sin5A + sin6A}{cos4A + cos5A + cos6A} = tan5A$

written 5.6 years ago by

teamques10 ★ 69k

modified 3.0 years ago by

pedsangini276 • 4.8k

engineering mathematics

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1 Answer

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written 5.6 years ago by

teamques10 ★ 69k

• modified 5.6 years ago

Solution:

L.H.S = $\frac{sin4A + sin5A + sin6A}{cos4A + cos5A + cos6A}$

$= \frac{sin4A + sin6A + sin5A}{cos4A + cos6A + cos5A}$

$= \frac{2sin\big(\frac{4A+6A}{2}\big)cos\big(\frac{4A-6A}{2}\big)+sin5A}{2cos\big(\frac{4A+6A}{2}\big)cos\big(\frac{4A-6A}{2}\big)+cos5A}$

$= \frac{2sin5Acos(-A)+sin5A}{2cos5Acos(-A)+ cos5A}$

$= \frac{sin5A(2cos(-A)+1)}{cos5A(2cos(-A)+1)}$

$= tan5A$

$=R.H.S$

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