Sin A -2 sin 3 A -2 cos 3 A -cos A -A- AB- AC- tan AD- 1

The correct option is C tan A sin A 2 sin^3 A/2 cos^3 A cos A = sin A (1 2 sin^2 A)/cos A (2 cos^2 A 1) = sin A (sin^2 A + cos^2 A 2 sin^2 A)/cos A(2 ...

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Standard X

Mathematics

Trigonometric Identities

sin A -2 sin ...

Question

$\frac{sin A - 2 {sin}^{3} A}{2 {cos}^{3} A - cos A} =$

sec A

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cot A

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tan A

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Solution

The correct option is C
tan A

$\frac{sin A - 2 {sin}^{3} A}{2 {cos}^{3} A - c o s A} = \frac{sin A (1 - 2 {sin}^{2} A)}{cos A (2 {cos}^{2} A - 1)}$

$= \frac{sin A ({sin}^{2} A + {cos}^{2} A - 2 {sin}^{2} A)}{cos A (2 {cos}^{2} A - ({sin}^{2} A + {cos}^{2} A))}$

$= \frac{sin A ({cos}^{2} A - {sin}^{2} A)}{cos A ({cos}^{2} A - {sin}^{2} A)}$

$= tan A$

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cos A - 2cos³ A/ 2sin³A - sin A = cot A

Q. Prove that

\frac{s i n A - 2 s i n^{3} A}{2 c o s^{3} A - c o s A} = t a n A

Q. Prove that:

\frac{sin A - 2 {sin}^{3} A}{2 {cos}^{3} A - cos A} = tan A

p r o v e t h a t \frac{s i n A - 2 {s i n}^{3} A}{2 {c o s}^{3} A - c o s A} = t a n A

Q. STATEMENT - 1 :

\frac{1 + s i n A}{1 - s i n A} = s e c A - t a n A

STATEMENT - 2 :

\frac{s i n A - 2 s i n^{2} A}{2 c o s^{3} A - c o s A} = t a n A

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sin A−2 sin3A2 cos3A−cos A=

The correct option is C tan A sin A−2 sin3A2 cos3A−cos A=sin A(1−2 sin2A)cos A(2 cos2A−1) =sin A(sin2A+cos2A−2 sin2A)cos A(2 cos2A−(sin2A+cos2A)) =sin A(cos2A−sin2A)cosA(cos2A−sin2A) =tan A

$\frac{sin A - 2 {sin}^{3} A}{2 {cos}^{3} A - cos A} =$