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Prove that $\frac{sin4A + sin5A + sin6A}{cos4A + cos5A + cos6A} = tan5A$
written 5.2 years ago by gravatar for teamques10 teamques10 ★ 68k modified 2.6 years ago by gravatar for pedsangini276 pedsangini276 • 4.8k
engineering mathematics
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written 5.2 years ago by gravatar for teamques10 teamques10 ★ 68k •   modified 5.2 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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