Question

please solve this problem in​

Answer 1

Answer:

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Step-by-step explanation:

$To \: prove = \\ \frac{tan \: θ + sec \: θ - 1}{tan \: θ - sec \: θ + 1} = \frac{1 + sin \:θ }{cos \: θ} \\ \\ so \: let \: \\ LHS = \frac{tan \: θ + sec \:θ - 1 }{tan \: θ - sec \: θ + 1} \\ \\ RHS = \frac{1 + sin \: θ}{cos \:θ }$

$so \: considering \: LHS \\ = \frac{tan \: θ + sec \:θ - 1 }{tan \:θ - sec \: θ + 1} \\ \\ so \: conjugate \: of \: tan \: θ - sec \: θ + 1 \: is \: tan \: θ + sec \: θ - 1 \\ so \: by \: conjugate \: method \\ we \: get \\ = \frac{tan \: θ + sec \: θ - 1}{tan \:θ - sec \:θ + 1 } \: \times \: \frac{tan \: θ + sec \:θ - 1 }{tan \: θ + sec \: θ - 1} \\ \\ = \frac{(tan \:θ + sec \: θ - 1) {}^{2} }{(tan \:θ - sec \: θ + 1)(tan \: θ + sec \:θ - 1) } \\ \\ so \: here \: (tan \: θ + sec \: θ - 1) {}^{2} \: is \: in \: form \: (a + b - c) {}^{2} \: and \\ (tan \: θ - sec \: θ + 1)(tan \: θ + sec \: θ - 1) \: is \: in \: from \: (a - b + c)(a + b - c) \\ so \: we \: know \: that \\ (a + b - c) {}^{2} = a { }^{2} + b {}^{2} + c {}^{2} + 2ab + - 2ac - 2bc \\ \\ (a - b + c)(a + b - c) = a {}^{2} - b {}^{2} - c {}^{2} + 2bc \\ \\ hence \: accordingly \\ = \frac{tan {}^{2} θ + sec {}^{2} θ + 1 + 2.tan \:θ.sec \:θ - 2.sec \: θ - 2.tan \: θ }{tan {}^{2} \: θ - sec {}^{2}θ - 1 + 2.sec \: θ} \\ \\ = \frac{sec {}^{2} \:θ + sec {}^{2} \: θ + 2.tan \: θ.sec \: θ - 2.tan \: θ - 2.sec \: θ}{sec {}^{2} \:θ - 1 - sec {}^{2} \:θ - 1 + 2.sec \: θ } \\ \\ since \: 1 + tan {}^{2} \: θ = sec {}^{2} θ$

$= \frac{2.sec {}^{2} \: θ \: + \: 2.tan \: θ.sec \:θ - 2.tan \: θ - 2.sec \:θ }{ - 2 + 2.sec \: θ} \\ \\ = \frac{2.sec {}^{2} \: θ - 2.sec \: θ + 2.tan \:θ.sec \: θ - 2.tan \: θ}{2(sec \:θ - 1) } \\ \\ = \frac{2.sec \: θ(sec \:θ - 1) + 2.tan \:θ(sec \: θ - 1) }{2(sec \:θ - 1) } \\ \\ = \frac{(2.sec \:θ + 2.tan \:θ)(sec \: θ - 1)}{ 2(sec \:θ - 1) } \\ \\ = \frac{2(sec \:θ + tan \: θ)(sec \: θ - 1)}{2(sec \: θ - 1)} \\ \\ = sec \: θ + tan \: θ$

$now \: we \: know \: that \\ sec \: θ = \frac{1}{cos \: θ} \\ \\ tan \: θ = \frac{sin \: θ}{cos \: θ} \\ \\ hence \: accordingly \\ \\ = \frac{1}{cos \:θ } + \frac{sin \:θ }{cos \: θ} \\ \\ = \frac{1 + sin \:θ }{cos \:θ } \\ \\ thus \: LHS=RHS \\ hence \: proved$