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,

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Solution

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

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

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

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

$= \frac{sin 4 A}{cos 4 A}$

$= tan 4 A = tan x$

$x = 4 A$

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\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 -

\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. 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) $\frac{s i n (θ + ϕ) - 2 s i n θ + s i n (θ + ϕ)}{c o s (θ + ϕ) - 2 c o s θ + c o s (θ + ϕ)}$

Q. Prove that

\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

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Prove: sinA+sin3A+sin5A+sin7AcosA+cos3A+cos5A+cos7A=tanxthen x is equal to,

The correct option is A 4AsinA+sin3A+sin5A+sin7AcosA+cos3A+cos5A+cos7A=tanx=(sinA+sin7A)+(sin3A+sin5A)(cosA+cos7A)+(cos3A+cos5A)=2sin4Acos3A+2sin4AcosA2cos4Acos3A+2cos4A+cosA=2sin4A(cos3A+cosA)2cos4A(cos3A+cosA)=sin4Acos4A=tan4A=tanxx=4A

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,