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

sin 4A

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cos 4A

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tan 4A

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cot 4A

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Solution

The correct option is D cot 4A
G.E= $\frac{(c o s A + c o s 5 A) + (c o s 3 A + c o s 7 A)}{(s i n A + s i n 5 A) + (s i n 3 A + s i n 7 A} = \frac{2 c o s 3 A c o s 2 A + 2 c o s 5 A c o s 2 A}{2 s i n 3 A c o s 2 A + 2 s i n 5 A c o s 2 A}$

$\frac{2 c o s 2 A (c o s 3 A + c o s 5 A)}{2 c o s 2 a (s i n 3 A + s i n 5 A)} = \frac{2 c o s 4 A c o s A}{2 s i n 4 A c o s A} = c o t 4 A$

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Similar questions

\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} =

\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 -

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

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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cosA+cos3A+cos5A+cos7AsinA+sin3A+sin5A+sin7A=

The correct option is D cot 4AG.E=(cosA+cos5A)+(cos3A+cos7A)(sinA+sin5A)+(sin3A+sin7A=2cos3A cos2A+2cos5A cos2A2sin3A cos2A+2sin5A cos2A 2cos2A(cos3A+cos 5A)2cos2a(sin3A+sin5A)=2cos 4AcosA2sin 4A cosA=cot 4A

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