$\frac{s i n A - 2 s i n^{3} A}{2 c o s^{3} A - c o s A}$

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tan A

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cot A

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sec A

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Solution

The correct option is B
tan A

$\frac{s i n A - 2 s i n^{3} A}{2 c o s^{3} A - c o s A}$

= $\frac{s i n A (1 - 2 s i n^{2} A)}{c o s A (2 c o s^{2} A - 1}$

= $\frac{s i n A (s i n^{2} A + c o s^{2} A - 2 s i n^{2} A)}{c o s A (2 c o s^{2} A - (s i n^{2} A + c o s^{2} A)}$

= $\frac{s i n A (c o s^{2} A - s i n^{2} A)}{c o s A (2 c o s^{2} A - s i n^{2} A - c o s^{2} A)}$

= $\frac{s i n A (c o s^{2} A - s i n^{2} A)}{c o s A (c o s^{2} A - s i n^{2} A)}$

= $\frac{s i n A}{c o s A}$

= tan A

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102

Similar questions

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

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

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

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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sinA−2sin3A2cos3A−cosA

The correct option is B tan A sinA−2sin3A2cos3A−cosA = sinA(1−2sin2A)cosA(2cos2A−1 = sinA(sin2A+cos2A−2sin2A)cosA(2cos2A−(sin2A+cos2A) = sinA(cos2A−sin2A)cosA(2cos2A−sin2A−cos2A) = sinA(cos2A−sin2A)cosA(cos2A−sin2A) = sinAcosA = tan A

$\frac{s i n A - 2 s i n^{3} A}{2 c o s^{3} A - c o s A}$