If sin A + cos A = m and

sec A + cosec A = n, show that :

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

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Solution

$s i n θ + c o s θ = m, s e c θ + c o s e c θ = n$
Consider LHS
$n (m^{2} - 1) = (s e c θ + c o s e c θ) [(c o s θ + s i n θ)^{2} - 1] = (s e c θ + c o s e c θ) [c o s^{2} θ + s i n^{2} θ + 2 s i n θ c o s θ - 1] = (s e c θ + c o s e c θ) [2 s i n θ c o s θ] s i n c e c o s^{2} θ + s i n^{2} θ = 1 = s e c θ . 2 s i n θ c o s θ + c o s e c θ . 2 s i n θ c o s θ = 2 s i n θ + 2 c o s θ = 2 [s i n θ + c o s θ] = 2 m$

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Similar questions

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

c o s e c A - sin A = m

sec A - cos A = n

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

\frac{cos A - sin A + 1}{cos A + sin A - 1} = cos e c A + cot A

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

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