Cos A-2 cos 3 A - 2 sin 3 A-sin A- A

Prove that (2cos3A cos A)/ (sin A – 2sin3A) = cot A We know cos2A = 1 sin2A So 2cos2A = 2 2sin2A L.H.S = (2cos3A cos A)/ (sin A – 2sin3A) = cos A*(2 ...

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Standard X

Mathematics

Standard Values of Trigonometric Ratios

cos A-2 cos 3...

Question

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

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Solution

Prove that

(2cos³A -cos A)/ (sin A – 2sin³A) = cot A

We know cos²A = 1- sin²A So 2cos²A = 2- 2sin²A

L.H.S = (2cos³A -cos A)/ (sin A – 2sin³A)

= cos A(2cos²A -1)/ [sin A (1 – 2sin²A)]

= cos A[( 2- 2sin²A)-1]/ [sin A (1 – 2sin²A)]

= cos A( 1- 2sin²A)/ sin A (1 – 2sin²A)

= cos A/SinA * [(1 – 2sin²A)/(1 – 2sin²A)]

= cos A/SinA * 1

= cot A

L.H.S = R.H.S

Hence Proved

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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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cos A - 2cos3 A/ 2sin3 A - sin A = cot A

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