Question

If sec A + tan A = x, obtain the values of sec A, tan A and sin A.

Answer 1

Answer:

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Step-by-step explanation:

$to \: find = \\ values \: of : - \\ 1.sec \: A \\ 2.tan \: A \\ 3.sin \: A \\ \\ given = \\ sec \: A + tan \: A = x \\ tan \: A = x - sec \: A \: \: \: \: \: (1) \\ \\ we \: know \: that \\ 1 + tan {}^{2} \: A = sec {}^{2} \: A \\ \\ 1 + (x - sec \: A) {}^{2} = sec {}^{2} \: A \\ 1 + x {}^{2} + sec {}^{2} \: A - 2.x.sec \: A = sec {}^{2} \: A \\ \\ 1 + x {}^{2} = 2.x.sec \: A \\ \\ sec \: A = \frac{1 + x {}^{2} }{2x} \\ \\ hence \: then \\ cos \: A = \frac{2x}{1 + x {}^{2} } \\ \\ now \: accordingly \\ \\ tan \: A = sec \: A - x \: \: \: \: \: from \: (1) \\ \\ = \frac{1 + x {}^{2} }{2x} - x \\ \\ = \frac{1 + x {}^{2} - 2x {}^{2} }{2x} \\ \\ tan \:A = \frac{1 - x {}^{2} }{2x}$

$so \: here \:as \\ \: cos \: A = \frac{2x}{1 + x {}^{2} } \\ \\ we \: know \: that \\ sin {}^{2} \: A + cos {}^{2} \: A = 1 \\ \\ sin {}^{2} \: A = 1 - cos {}^{2} A \\ \\ = 1 - ( \frac{2x}{1 + x {}^{2} } \: ) {}^{2} \\ \\ = 1 - \frac{4x {}^{2} }{1 + x {}^{2} + 2x {}^{2} } \\ \\ = \frac{1 + x {}^{4} +2x {}^{2} - 4x {}^{2} }{1 + x {}^{4} + 2x {}^{2} } \\ \\ = \frac{1 + x {}^{4} - 2x {}^{2} }{1 + x {}^{4} + 2x {}^{2} } \\ \\ = \frac{(1 - x {}^{2}) {}^{2} }{(1 + x {}^{2}) {}^{2} } \: \\ \\ = ( \frac{1 - x {}^{2} }{1 + x {}^{2} } ) {}^{2} \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: get \\ \\ sin \: A =± \: \frac{1 - x {}^{2} }{1 + x {}^{2} }$