Question

Prove that :-

sec θ ( 1 - sin θ ) ( sec θ + tan θ ) = 1​

Answer 1

L.H.S =

$\bf \implies{ sec \: θ ( 1 - sin \: θ ) ( sec \: θ + tan \: θ ) = 1} \: \: \: \: \\ \\ \bf \implies = \dfrac {1}{cos \: θ} \: \: (1 - sin \: θ) \: \times \: ( \dfrac {1}{cos \: θ} \: + \: \dfrac{sin \: θ}{cos \:θ})$

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

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

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

$\\ \bf \implies = 1 \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bf \implies = R.H.S$

$\\ \bf{ \: ∴ \: sec \: θ ( 1 - sin \: θ ) \: ( sec \: θ + tan \: θ ) = 1}$

Answer 2

Answer:

L.H.S. = sec θ (1 – sin θ) (sec θ + tan θ) = 1 = R.H.S. ∴ sec θ (1 – sin θ) (sec θ + tan θ) = 1 ii. L.H.S. = (sec θ + tan θ) (1 – sin θ) = cos θ = R.H.S. ∴ (sec θ + tan θ) (1 – sin θ) = cos θRead more on Sarthaks.com - https://www.sarthaks.com/857818/prove-the-following-i-sec-1-sin-sec-tan-1-ii-sec-tan-1-sin-cos

sec θ ( 1 - sin θ ) ( sec θ + tan θ ) =