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Trigonometric Ratios of Multiples of an Angle

If m∧ 2+m'∧ 2...

Question

If m^2 + m'^ 2 + 2mm' cos theta = 1 , n^2+ n' ^2 + 2nn' cos theta = 1 and mn + m'n' + (mn' + m'n) cos theta = 0, prove that m^2 + n^2 = cosec^2 theta .

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Solution

m’² + m² + 2mm’cosθ = 1....(X)

Spitting m² as m²(sin²θ + m²cos²θ)

m²= m²sin²θ + m²cos²θ

Therefore (X) can be written as

m’² + m²sin²θ + m²cos²θ + 2mm′cosθ = 1

m’² + m²cos²θ + 2mm’cosθ = 1 - m²sin²θ

=> (m’ + mcosθ)² = 1 -m²sin²θ

Similarly for n and n'

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Similar questions

m^{2} + {m^{'}}^{2} + 2 m m^{'} cos θ = 1

n^{2} + {n^{'}}^{2} + 2 n n^{'} cos θ = 1

m n + m^{'} n^{'} + (m n^{'} + m^{'} n) = cos θ = θ

, then
prove that

m^{2} + n^{2} = {csc}^{2} θ

m^{2} + {m^{'}}^{2} + 2 m m^{'} cos x = 1

n^{2} + {n^{'}}^{2} + 2 n n^{'} cos x = 1

m n + m^{'} n^{'} + (m n^{'} + m^{'} n) cos x = 0

, prove that
(i)

m^{2} + n^{2} = c o {sec}^{2} x

{m^{'}}^{2} + {n^{'}}^{2} = c o {sec}^{2} x

c o s e c θ - sin θ = m

sec θ - cos θ = n

(m^{2} n)^{2 / 3} + (m n^{2})^{2 / 3} = 1

If $(c o t θ + t a n θ) = m$ and $(s e c θ - c o s θ) = n$ , prove that

$(m^{2} n)^{\frac{2}{3}} - (m n^{2})^{\frac{2}{3}} = 1$

cos θ > 0, tan θ + sin θ = m

tan θ - sin θ = n

, then show that

m^{2} - n^{2} = 4 \sqrt{m n}

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