Question

please solve this problem

Answer 1

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Answer 2

$\bold{\underline{\underline{\huge{Solution}}}}$

$\implies \dfrac{sec\: \theta \: + \: tan\: \theta \: - \: ( sec^{2}\: \theta \: - \: tan^{2}\: \theta)}{(tan\: \theta \: - \: sec\: \theta \: + \: 1)}$

$\implies \dfrac{(sec\: \theta \: + \: tan\: \theta)\: - \: [(sec\: \theta \: + \: tan\: \theta)(sec\: \theta \: - \: tan\: \theta)]}{(tan\: \theta \: - \: sec\: \theta\: + \: 1)}$

$\implies (sec\: \theta \: + \: tan\: \theta)\dfrac{[1\: - \: (sec\: \theta\: - \: tan \: \theta)]}{tan\: \theta \: - \: sec\: \theta \: + \: 1}$

$\implies (sec\: \theta \: + \: tan\: \theta)\dfrac{(1\: - \: sec\: \theta\: + \: tan \: \theta)}{tan\: \theta \: - \: sec\: \theta \: + \: 1}$

$\implies (sec\: \theta \: + \: tan\: \theta)\dfrac{(1\: - \: sec\: \theta\: + \: tan \: \theta\: )}{(1 \: - \: sec\: \theta \: + \: tan\: \theta)}$

after elimination :

$\therefore\: ( sec \: \theta\: + \: tan \: \theta )$

$\implies \dfrac{1}{cos\: \theta} \: + \: \dfrac{sin\: \theta}{cos\: \theta}$

$\implies \dfrac{1\: + \: sin\: \theta}{cos\: \theta}$

multiply numerator and denominator by ( 1 - sin @ )

$\implies \dfrac{1\: + \: sin\: \theta}{cos\: \theta}\times\dfrac{1\: - \: sin \: \theta}{1\: - \: sin \: \theta}$

$\implies \dfrac{1 \: - \: sin^{2}\: \theta}{cos\: \theta \: ( 1 \: - \: sin\: \theta)}$

$\implies \dfrac{cos^{2}\: \theta}{cos\: \theta \: ( 1 \: - \: sin\: \theta)}$

$\implies \dfrac{cos\: \theta}{( 1 \: - \: sin\: \theta)}$

RHS

$\bold{\boxed{\boxed{LHS \: = \: RHS}}}$

$\bold{\huge{Hence \: Proved}}$