Question

prove the following

Answer 1

\[1 - \sin \theta = \frac{ \cos ^2 \theta }{ 1 + \sin \theta }\]

Answer 2

LHS= (1-sinx) =(1-sinx)*(1+sinx) / (1+sinx)
= (1-(sinx)^2) / (1+sinx)
=cos^2 x/ (1+sinx) [as 1-sin^2x =cos^2x ]

Answer 3

alternatively, you can use that fact that cos^2(theta) = 1-sin^2(theta)

then factor using the difference of squares formula

Answer 4

so what do i do?

Answer 5

\[\Large 1 - \sin \theta = \frac{ \cos ^2 \theta }{ 1 + \sin \theta }\]
\[\Large 1 - \sin \theta = \frac{ 1-\sin ^2 \theta }{ 1 + \sin \theta }\]

what's next?

Answer 6

i have no idea

Answer 7

factor the numerator on the right side using the difference of squares formula

Answer 8

difference of squares formula: a^2 - b^2 = (a-b)(a+b)

Answer 9

i have no idea how to do this problem

Answer 10

let's say we had 1 - x^2

how would you factor that using the difference of squares formula