$\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A}$ is equal to -

sin 4 A

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cos 4 A

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tan 4 A

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None of these

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Solution

The correct option is B $tan 4 A$
$\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A}$

$= \frac{(sin A + sin 7 A) + (sin 5 A + sin 3 A)}{(cos A + cos 7 A) + (cos 5 A + cos 3 A)}$

$= \frac{(2 sin 4 A . cos 3 A) + (2 sin 4 A . cos A)}{(2 cos 4 A . cos 3 A) + (2 cos 4 A . cos A)}$

$= \frac{sin 4 A (2 cos 3 A + 2 cos A)}{cos 4 A (2 cos 3 A + 2 cos A)} = tan 4 A$

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Similar questions

Q. Prove that

\frac{sinA + s i n 3 A + s i n 5 A + s i n 7 A}{c o s A + cos 3 A + cos 5 A + cos 7 A} = tan 4 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} =

Q. (i)

\frac{cos 3 θ + 2 cos 5 θ + cos 7 θ}{cos θ + 2 cos 3 θ + cos 5 θ} = cos 2 θ - sin 2 θ tan 3 θ

(ii)

\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A} = tan 4 A

Q. Solve the problem:-

\frac{sin A - sin 3 A + sin 5 A - sin 7 A}{cos A - cos 3 A - cos 5 A + cos 7 A} = 2 cot 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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sinA+sin3A+sin5A+sin7AcosA+cos3A+cos5A+cos7A is equal to -

The correct option is B tan4A sinA+sin3A+sin5A+sin7AcosA+cos3A+cos5A+cos7A=(sinA+sin7A)+(sin5A+sin3A)(cosA+cos7A)+(cos5A+cos3A)=(2sin4A.cos3A)+(2sin4A.cosA)(2cos4A.cos3A)+(2cos4A.cosA)=sin4A(2cos3A+2cosA)cos4A(2cos3A+2cosA)=tan4A

$\frac{sin A + sin 3 A + sin 5 A + sin 7 A}{cos A + cos 3 A + cos 5 A + cos 7 A}$ is equal to -