Question

Prove that :( secθ + tanθ) (1 - sinθ) = cosθ​

Answer 1

$\bf \large \implies{ \green{L.H.S}= (sec \: \theta + tan \: \theta ) \: (1 - sin \: \theta)}$

$\bf \large\implies{ = \Huge [ } \: \large{\dfrac{1}{cos \: \theta} + \dfrac{sin \: \theta}{cos \: \theta}} \: {\Huge]} \bf \large{(1 - sin \: \theta)}$

$\bf \large \implies = \dfrac{(1 + sin \: \theta)}{cos \: \theta} \: (1 - \: sin \: \theta)$

$\bf \large \implies{ = \dfrac{1 - {sin}^{2} \: \theta}{cos \: \theta} } \: \: \: \: \purple{ ... \: [\: (a + b)(a - b) = {a}^{2} \: - \: {b}^{2}]}$

$\bf \large\implies = \dfrac{ {cos}^{2} \: \theta}{ {cos}^{2} \: \theta} \: \: \: \: \: \: \: \: \: \: \: \: \large \binom{ { sin}^{2} \: \theta + {cos}^{2} \: \theta = 1 \: }{ 1 - {sin}^{2} \: \theta = {cos} \: \theta\: }$

$\bf \large \implies{ = cos \: \theta}$

$\bf \large \implies = \green {{ R.H.S}}$

$\bf \large \red{\therefore(sec \: \theta \: + \: tan \: \theta)(1 - sin \: \theta) = cos \: \theta}$