Question

(Sec A + tan A) (1-sinA) =​

Answer 1

$\bf\huge \gray{ANSWER}$

$\large \text{(sec \: A+tan \: A)(1−sin \: A) = cos \: A }$

$\large \text\blue{Explanation}$

$\large \text{L.H S =(Sec A + tan A) (1-sinA) }$

∴

$\sf \large = > { (\frac{1}{cos \: A} + \frac{ sin \:A}{cos \:A }) \: (1-sin \: A)} \: >> [(sec \: A = \frac{1}{cos \: A} )tan \: A = \frac{sin \:A }{cos \: A} ]$

∴

$\sf{ = > ( \frac{(1 + sin \:A)(1 - sin \: A) }{cos \: A})(1 - sin \: A) }$

∴

$\sf{ = = > \frac{1+(sinA)(1−sinA)}{cos \:A } = }$

∴

$\sf{ = = > \frac{(1 - {sin}^{2}A)}{cos \: A}}$

∴

$\sf{[Formula \: used =(a+b)(a-b)= (a²+b²)]}$

∴

$\sf{=> \frac{ {cos}^{2}\:A}{cos \:A}}= cos \:A$

∴

$\bf\huge \gray{Therefore}$

$\bf{∴(sec \: A+tan \: A)(1−sin \: A)=cos \: A}$