Question

prove the following fast pls...​

Answer 1

$1) \: (1 - { \sin }^{2} A) { \sec }^{2} A = 1$

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

$2) \: { \sec }^{4} - { \sec }^{2} \theta = { \tan }^{4} \theta + { \tan }^{2}$

$= > \: { \sec }^{4} - { \sec }^{2} \theta = {sec}^{2} \theta( { \sec }^{2} \theta - 1)$

$= (1 + { \tan }^{2} \theta)( { \tan }^{2} \theta)$

$= { \tan }^{2} \theta + { \tan }^{4} \theta$

$3) (\ \sec \theta - \tan \theta {)}^{2} = \frac{1 - \sin \theta }{1 + \sin \theta }$

$= > ( \sec \theta - \tan \theta {)}^{2} = \frac{(1 - \sin \theta {)}^{2} }{ { \cos }^{2} \theta }$

$= \frac{(1 - \sin \theta {)}^{2} }{1 - { \sin}^{2} \theta}$

$= \frac{1 - \sin \theta }{1 + \sin \theta }$

4) is in the attachment......