Solve- sin A - sin 5A - sin 9A - sin 13Acos A - cos 5A - cos 9A - cos 13A - 4A

LHS = sin A sin 5A+sin A sin 13A/cos A cos 5A cos 9A+cos 13A #160; =(sin A sin 13A) (sin 5A sin 9A)/(cos A+cos 13A)(cos 5A+cos 9A) #160;we know, ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Allied Angles

Solve: sin A...

Question

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

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Solution

LHS = $\frac{sin A - sin 5 A + sin A - sin 13 A}{cos A - cos 5 A - cos 9 A + cos 13 A}$

$= \frac{(sin A - sin 13 A) - (sin 5 A - sin 9 A)}{(cos A + cos 13 A) (cos 5 A + cos 9 A)}$

we know,
$sin A - sin B = 2 cos (\frac{A + B}{2}) sin (\frac{A - B}{2})$

$cos A + cos B = 2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})$

$∴ L H S = \frac{2 cos (\frac{13 A + A}{2}) sin (\frac{A - 13 A}{2}) - 2 cos (\frac{9 A + 5 A}{2}) sin (\frac{5 A - 9 A}{2})}{2 cos (\frac{A + 13 A}{2}) cos (\frac{A - 13 A}{2}) 2 c o s (\frac{5 A + 9 A}{2}) cos (\frac{5 A - 9 A}{2})}$

$= \frac{2 cos 7 A sin (- 6 A) - 2 cos 7 A sin (- 2 A)}{2 cos 7 A cos (- 6 A) - 2 cos 7 A cos (- 2 A)}$

$= \frac{2 cos 7 A (sin 2 A - sin 6 A)}{2 cos 7 A (cos 6 A - cos 2 A)}$

$= \frac{2 cos (\frac{6 A + 2 A}{2}) sin (\frac{2 A + 6 A}{2})}{- 2 sin (\frac{6 A + 2 A}{2}) sin (\frac{6 A - 2 A}{2})}$

$= \frac{- cos 4 A sin 2 A}{- sin 4 A}$
$= cot 4 A$

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Similar questions

Q. 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

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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Solve:sinA−sin5A+sin9A−sin13AcosA−cos5A−cos9A+cos13A=cot4A

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