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Trigonometric Identities

If tan θ+ sin...

Question

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

$(m^{2} - n^{2})^{2} = 16 m n .$

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Solution

$m = (t a n θ + s i n θ) n = (t a n θ - s i n θ) m^{2} - n^{2} = (m + n) (m - n) = (t a n θ + s i n θ + t a n θ - s i n θ) (t a n θ + s i n θ - (t a n θ - s i n θ)) = (2 t a n θ) (t a n θ + s i n θ - t a n θ + s i n θ) = (2 t a n θ) (2 s i n θ) = 4 t a n θ s i n θ (m^{2} - n^{2})^{2} = 16 s i n^{2} θ t a n^{2} θ - - - (1) m n = (t a n θ + s i n θ) (t a n θ - s i n θ) = t a n^{2} θ - s i n^{2} θ = s i n^{2} θ (\frac{1}{c o s^{2} θ} - 1) = s i n^{2} θ (\frac{1 - c o s^{2} θ}{c o s^{2} θ}) = s i n^{2} θ (\frac{s i n^{2} θ}{c o s^{2} θ}) = s i n^{2} θ t a n^{2} θ 16 m n = 16 s i n^{2} θ t a n^{2} θ - - - - (2) From (1) and (2), (m^{2} - n^{2})^{2} = 16 m n$

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Similar questions

tan θ + sin θ = m

tan θ - sin θ = n

[m^{2} - n^{2}]^{2} = 16 m n

tan θ + sin θ = m, tan θ - sin θ = n

(m^{2} - n^{2})^{2} =

.

^{'} θ^{'}

tan θ + sin θ = m, tan θ - sin θ = n

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