Find the value ofsin 5A-sin 3Acos 5A-cos 3A

Taking LHS, sin 5A sin 3Acos 5A+cos 3A nbsp; nbsp; nbsp;=2cos ( 5A+3A2 ).sin ( 5A 3A2 )2cos ( 5A+3A2 ).cos ( 5A 3A2 ) Using nbsp;sin C sin D=2cos ...

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Standard X

Mathematics

Trigonometric Ratios

Find the valu...

Question

Find the value of
$\frac{sin 5 A - sin 3 A}{cos 5 A + cos 3 A}$

sin A

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tan A

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Solution

The correct option is B tan A
Taking LHS,

$\frac{sin 5 A - sin 3 A}{cos 5 A + cos 3 A}$

$= \frac{2 cos (\frac{5 A + 3 A}{2}) . sin (\frac{5 A - 3 A}{2})}{2 cos (\frac{5 A + 3 A}{2}) . cos (\frac{5 A - 3 A}{2})}$

Using $sin C - sin D = 2 cos (\frac{C + D}{2}) . sin (\frac{C - D}{2}) a n d cos C + cos D = 2 cos (\frac{C + D}{2}) . sin (\frac{C - D}{2})$

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

= tan A = RHS

Hence, proved

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Similar questions

Find the value of $\frac{c o s^{3} A - c o s 3 A}{c o s A}$ + $\frac{s i n^{3} A - s i n 3 A}{s i n A}$

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

\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. 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

\frac{s i n^{3} A + s i n 3 A}{s i n A} + \frac{c o s^{3} A - c o s 3 A}{c o s A}

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Find the value of sin5A−sin3Acos5A+cos3A

The correct option is B tan ATaking LHS, sin5A−sin3Acos5A+cos3A =2cos(5A+3A2).sin(5A−3A2)2cos(5A+3A2).cos(5A−3A2) Using sinC−sinD=2cos(C+D2).sin(C−D2) and cosC+cosD=2cos(C+D2).sin(C−D2) =sinAcosA = tan A = RHS Hence, proved

Find the value of
$\frac{sin 5 A - sin 3 A}{cos 5 A + cos 3 A}$