Prove that sin A - sin 5 A - sin 9 A cos A - cos 5 A - cos 9 A - tan 5 A

sin A +Sin 5A +sin 9Acos A+cos 5A+cos 9A=tan 5A LHS. =sin A +sin 5A+sin 9Acos A+cos 5A+cos 9A=sin (5A 4A)+sin 5A+sin (5A+4Acos(5A 4A)+cos 5A+cos (5A+4A ...

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Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

Prove that ...

Question

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

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Solution

$\frac{sin A + S i n 5 A + sin 9 A}{cos A + cos 5 A + cos 9 A} = tan 5 A$
LHS.
$= \frac{sin A + sin 5 A + sin 9 A}{cos A + cos 5 A + cos 9 A} = \frac{sin (5 A - 4 A) + sin 5 A + sin (5 A + 4 A}{cos (5 A - 4 A) + cos 5 A + cos (5 A + 4 A)}$
$= \frac{sin 5 A . cos 4 A - sin 4 A . cos 5 A + sin 5 A + s i n 5 A . c o s 4 A + sin 4 A . cos 5 A}{cos 5 A . cos 4 A + sin 5 A . sin 4 A + cos 5 A + cos 5 A . cos 4 A - sin 5 A . sin 4 A} .$
$= \frac{sin 5 A [2 cos 4 A + 1]}{cos 5 A [2 cos 4 A + 1]} = \frac{sin 5 A}{cos 5 A} = tan 5 A$

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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. 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:

\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A} = tan x

then

x

is equal to,

Q. Prove that:
(i)

\frac{\sin A + \sin 3 A + \sin 5 A}{\cos A + \cos 3 A + \cos 5 A} = \tan 3 A

(ii)

\frac{\cos 3 A + 2 \cos 5 A + \cos 7 A}{\cos A + 2 \cos 3 A + \cos 5 A} = \frac{\cos 5 A}{\cos 3 A}

(iii)

\frac{\cos 4 A + \cos 3 A + \cos 2 A}{\sin 4 A + \sin 3 A + \sin 2 A} = \cot 3 A

(iv)

\frac{\sin 3 A + \sin 5 A + \sin 7 A + \sin 9 A}{\cos 3 A + \cos 5 A + \cos 7 A + \cos 9 A} = \tan 6 A

(v)

\frac{\sin 5 A - \sin 7 A + \sin 8 A - \sin 4 A}{\cos 4 A + \cos 7 A - \cos 5 A - \cos 8 A} = \cot 6 A

(vi)

\frac{\sin 5 A \cos 2 A - \sin 6 A \cos A}{\sin A \sin 2 A - \cos 2 A \cos 3 A} = \tan A

(vii)

\frac{\sin 11 A \sin A + \sin 7 A \sin 3 A}{\cos 11 A \sin A + \cos 7 A \sin 3 A} = \tan 8 A

(viii)

\frac{\sin 3 A \cos 4 A - \sin A \cos 2 A}{\sin 4 A \sin A + \cos 6 A \cos A} = \tan 2 A

(ix)

\frac{\sin A \sin 2 A + \sin 3 A \sin 6 A}{\sin A \cos 2 A + \sin 3 A \cos 6 A} = \tan 5 A

(x)

\frac{\sin A + 2 \sin 3 A + \sin 5 A}{\sin 3 A + 2 \sin 5 A + \sin 7 A} = \frac{\sin 3 A}{\sin 5 A}

(xi)

\frac{\sin (θ + ϕ) - 2 \sin θ + \sin (θ - ϕ)}{\cos (θ + ϕ) - 2 \cos θ + \cos (θ - ϕ)} = \tan θ

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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Prove that sinA+sin5A+sin9AcosA+cos5A+cos9A=tan5A

sinA+Sin5A+sin9AcosA+cos5A+cos9A=tan5ALHS.=sinA+sin5A+sin9AcosA+cos5A+cos9A=sin(5A−4A)+sin5A+sin(5A+4Acos(5A−4A)+cos5A+cos(5A+4A)=sin5A.cos4A−sin4A.cos5A+sin5A+sin5A.cos4A+sin4A.cos5Acos5A.cos4A+sin5A.sin4A+cos5A+cos5A.cos4A−sin5A.sin4A.=sin5A[2cos4A+1]cos5A[2cos4A+1]=sin5Acos5A=tan5A

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