Question

prove (1-sin x)/sec x = cos^3 x/1+sin x
any trig people in the houseeeeee

Answer 1

i started by multiplying the left side by the conjugate. but i know you can factor the other side too and start that way

Answer 2

With trig id problems, its helpful to start working on a side that has a lot of algebra that can be done. for example, I'll get your started

\[ \frac{1-\sin x}{\sec x} = \frac{\cos^3 x}{1+\sin x} \] It seems that the right hand side has a lot of manipulation potential. we can:
\[ \frac{1-\sin x}{\sec x} = \frac{(\cos^2 x)(\cos x)}{1+\sin x} \]
Recall
\[\cos^2x = 1-\sin^2 x \] Then
\[ \frac{1-\sin x}{\sec x} = \frac{(1-\sin^2x)(\cos x)}{1+\sin x} \]
Maybe you can multipled the right hand side by the conjugate to keep it going.

Answer 3

In fact, if you do, you may almost be done.

Answer 4

no i dont know what im doing

Answer 5

\[\huge \frac{ (1-\sin x)}{\sec x}\]
multiplying Nr. and Dr. by (1+ sin x) we find
\[\huge \frac{ (1-\sin x)}{\sec x} = \frac{ (1-\sin x)}{\sec x} \times \frac{ (1+\sin x)}{(1+\sin x)}\]
\[\huge = \frac{ (1^2-\sin^2 x)}{\frac{1}{\cos x}(1+\sin x)}\]
\[\huge = \frac{ \cos^2 x \times \cos x}{(1+\sin x)} = \frac{\cos^3x}{ 1 +sinx}\]

Answer 6

I GET ITTTTTTTTTTTTTTTTTTTTTTT

Answer 7

if you dont know those identities youre done. you fail.

Answer 8

what happened with 1/cos x(1+sinx) for the demoninators

Answer 9

\[\sec x = \frac{1}{cosx} \] then simply by multiplying by the reciprocal

Answer 10

how did that become 1+sinx

Answer 11

The (1 + sin(x) stayed but the 1/cos(x) was flipped and then multiplied to the fraction (that's where the 2nd cos came from in the numerator up there)

Answer 12

\[\frac{\cos^2 x}{\frac{1}{\cos x}(1+\sin x)} = \frac{\cos^2 x}{(1+\sin x)} \frac{\cos x}{1}\]

Answer 13

\[\huge \frac{ (1^2-\sin^2 x)}{\frac{1}{\cos x}(1+\sin x)} = \huge \frac{ (1^2-\sin^2 x)}{(1+\sin x)} \times cos(x)\]

Answer 14

okay all of you have helped me so much i wish i could medal all of you