The value of sin 5 -cos 3 -cos 5 -2 cos 4 -cos 3 - is

The correct option is D tan α/2 sin 5 α sin 3 α/cos 5 α + 2 cos 4 α + cos 3 α = sin 5 α sin 3 α/cos 5 α + cos 3 α + 2 cos 4 α = 2 sin α cos 4 α/2 cos 4 α ...

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Standard XII

Mathematics

Basic Trigonometric Identities

The value of ...

Question

The value of $\frac{s i n 5 α - s i n 3 α}{c o s 5 α + 2 c o s 4 α + c o s 3 α}$ is

$c o t \frac{α}{2}$

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$c o t α$

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$t a n \frac{α}{2}$

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none of these

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Solution

The correct option is C
$t a n \frac{α}{2}$

$\frac{s i n 5 α - s i n 3 α}{c o s 5 α + 2 c o s 4 α + c o s 3 α} = \frac{s i n 5 α - s i n 3 α}{c o s 5 α + c o s 3 α + 2 c o s 4 α} = \frac{2 s i n α c o s 4 α}{2 c o s 4 α c o s α + 2 c o s 4 α} = \frac{2 s i n α c o s 4 α}{2 c o s 4 α (c o s α + 1)} = \frac{s i n α}{c o s α + 1}$

$= \frac{2 s i n \frac{α}{2} . c o s \frac{α}{2}}{c o s^{2} \frac{α}{2} - s i n^{2} \frac{α}{2} + s i n^{2} \frac{α}{2} + c o s^{2} \frac{α}{2}} = \frac{2 s i n \frac{α}{2} c o s \frac{α}{2}}{2 c o s^{2} \frac{α}{2}} = \frac{s i n \frac{α}{2}}{c o s \frac{α}{2}} = t a n \frac{α}{2}$

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Similar questions

Q. The value of

\frac{\sin 5 α - \sin 3 α}{\cos 5 α + 2 \cos 4 α + \cos 3 α} =

(a)

\cot α / 2

(b)

\cot α

(c)

\tan α / 2

(d) None of these

Q. If

f_{n} (α) = \frac{sin α + sin 3 α + sin 5 α + . . . . . + sin (2 n - 1) α}{cos α + cos 3 α + cos 5 α + . . . . . + cos (2 n - 1) α}

find

f_{4} (π / 32)

Q. Prove the following identity:

\frac{{sin}^{2} 4 α}{2 cos α + cos 3 α + cos 5 α} = 2 sin α sin 2 α

Q. Assertion :

{cos}^{3} α + {cos}^{3} (α + \frac{2 π}{3}) + {cos}^{3} (α + \frac{4 π}{3}) = 3 cos α cos (α + \frac{2 π}{3}) cos (α + \frac{4 π}{3})

Reason: If

a + b + c = 0 \Leftrightarrow a^{3} + b^{3} + c^{3} = 3 a b c

Q. If

tan α = 2

, then the value of

\frac{sin α}{{sin}^{3} α + {cos}^{3} α}

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The value ofsin 5α−sin 3αcos 5α+2 cos 4α+cos 3α is

The value of $\frac{s i n 5 α - s i n 3 α}{c o s 5 α + 2 c o s 4 α + c o s 3 α}$ is