Show that- sinA - sin5A - sin9AcosA - cos5A - cos9A -tan5A

L.H.S = sinA+sin5A+sin9A/cosA+cos5A+cos9A = (sin9A+sinA)+sin5A/(cosA+coa9A)+cos5A = 2sin(A+9A/2)cos(9A A/2)+sin5A/2cos(A+9A/2). cos(9A A/2)+cos5A ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Ratios

Show that: s...

Question

Show that:
$\frac{s i n A + s i n 5 A + s i n 9 A}{c o s A + c o s 5 A + c o s 9 A} = t a n 5 A$

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Solution

L.H.S = $\frac{s i n A + s i n 5 A + s i n 9 A}{c o s A + c o s 5 A + c o s 9 A}$
$= \frac{(s i n 9 A + s i n A) + s i n 5 A}{(c o s A + c o a 9 A) + c o s 5 A}$
$= \frac{2 s i n (\frac{A + 9 A}{2}) c o s (\frac{9 A - A}{2}) + s i n 5 A}{2 c o s (\frac{A + 9 A}{2}) . c o s (\frac{9 A - A}{2}) + c o s 5 A}$
$= \frac{2 s i n 5 A c o s 4 A + s i n 5 A}{2 c o s 5 A c o s 4 A + c o s 5 A}$
$= \frac{s i n 5 A (2 c o s 4 A + 1)}{c o s 5 A (2 c o s 4 A + 1)}$
= tan 5A
= R.H.S

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Similar questions

Q. Solve:

\frac{sin A - sin 5 A + sin 9 A - sin 13 A}{cos A - cos 5 A - cos 9 A + cos 13 A} = cot 4 A

Q. Prove that

\frac{sin A + sin 5 A + sin 9 A}{cos A + cos 5 A + cos 9 A} = tan 5 A

\frac{sin A + sin 2 A + sin 4 A + sin 5 A}{cos A + cos 2 A + cos 4 A + cos 5 A} =

\frac{c o s A + c o s 3 A + c o s 5 A + c o s 7 A}{s i n A + s i n 3 A + s i n 5 A + s i n 7 A} =

8. Prove that :
(i) $\frac{s i n A + s i n 3 A + s i n 5 A}{c o s A + c o s 3 A + c o s 5 A} = t a n 3 A$
(ii) $(i i) \frac{c o s 3 A + 2 c o s 5 A + c o s 7 A}{c o s A + 2 c o s 3 A + c o s 5 A} = \frac{c o s 5 A}{c o s 3 A}$
(iii) $\frac{c o s 4 A + c o s 3 A + c o s 2 A}{c o s 4 A + s i n 3 A + s i n 2 A} = c o t 3 A$
(iv) $\frac{s i n 3 A + s i n 5 A + s i n 7 A + s i n 9 A}{c o s 3 A + c o s 5 A + c o s 7 A + c o s 9 A} = t a n 6 A$
(v) $\frac{s i n 5 A - s i n 7 A + s i n 8 A - s i n 4 A}{c o s 4 A + c o s 7 A + c o s 7 A - c o s 5 A - c o s 8 A} = c o t 6 A$
(vi) $\frac{s i n 5 A + c o s 2 A - s i n 6 A c o s A}{s i n A s i n 2 A - c o s 2 A c o s 3 A} = t a n A$
(vii) $\frac{s i n 11 A + s i n A + s i n 7 A + s i n 3 A}{c o s 11 A s i n A + c o s 7 A s i n 3 A} = t a n 8 A$
(viii) $\frac{s i n 3 A c o s 4 A - s i n A c o s 2 A}{s i n 4 A s i n A + c o s 6 A c o s A} = t a n 2 A$
(ix) $\frac{s i n A s i n 2 A + s i n 3 A s i n 6 A}{s i n A c o s 2 A + c o s 3 A c o s 6 A} = t a n 5 A$
(x) $\frac{s i n A + 2 s i n 3 A + s i n 5 A}{s i n 3 A + 2 s i n 5 A + s i n 7 A} = \frac{s i n 3 A}{s i n 5 A}$
(xi) $\frac{s i n (θ + ϕ) - 2 s i n θ + s i n (θ + ϕ)}{c o s (θ + ϕ) - 2 c o s θ + c o s (θ + ϕ)}$

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Show that:sinA+sin5A+sin9AcosA+cos5A+cos9A=tan5A

L.H.S = sinA+sin5A+sin9AcosA+cos5A+cos9A=(sin9A+sinA)+sin5A(cosA+coa9A)+cos5A=2sin(A+9A2)cos(9A−A2)+sin5A2cos(A+9A2).cos(9A−A2)+cos5A=2sin5Acos4A+sin5A2cos5Acos4A+cos5A=sin5A(2cos4A+1)cos5A(2cos4A+1)= tan 5A= R.H.S

Show that:
$\frac{s i n A + s i n 5 A + s i n 9 A}{c o s A + c o s 5 A + c o s 9 A} = t a n 5 A$