stuck on trig problem

[yasaminG]

so the original problem was

(sec x - tan x ) ^2 = 1- sin x / 1+ sin x

we are suppose to prove that the left is equal to the right ;however , i am totally stuck on the last few steps. can some please help me!

thank you

 

[yasaminG]

please help me!!

im up to where

=1- (2sinx+sin^2x) / cos^2x

 

[Daniel_Feldman]

This problem is much easier to do backwards (which is allowed!).

In other words, it's easier to start with 1- sin x / 1+ sin x and make it equal the other side.

1- sin x / 1+ sin x multiplied by (1-sinx)/(1-sinx)


=(1-2sinx+sin^2(x))/(1-sin^2(x))

Now, remember the identity:

sin^2(x)+cos^2(x)=1

so

1-sin^2(x)=cos^2(x)


Thus, we can rewrite what we have so far.

(1-2sinx+sin^2(x))/(1-sin^2(x))=(1-2sin(x)+sin^2(x))/(cos^2(x))

=(1/cos^2(x))-(2sin(x)/cos^2(x))+(sin^2(x)/cos^2(x))

=sec^2(x)-2tan(x)sec(x)+tan^2(x)

Factoring yields

(secx-tanx)^2

 

[yasaminG]

yea i could have done that, but i wanted to start with the harder side,
so ill try doing it your way but can you see why im stuck?

 

[Guest]

taking the right side to start with...

= ( (1-sinx) . (1-sinx) )/ ( (1+sinx) (1-sinx) )

= ( 1- 2 sinx + (sinx)^2 ) / ( 1 - (sinx)^2 )

= ( 1- 2 sinx + (sinx)^2 ) / ( (cos x)^2 )

now write each of the items over the ( ( cosx)^2) and simply and ok

 

[yasaminG]

yea i could have done that, but i wanted to start with the harder side,
so ill try doing it your way but can you see why im stuck?

 

[soroban]

Hello, yasaminG!

Prove: .$\displaystyle (\sec x\,-\,\tan x)^2\:=\:\frac{1\,-\,\sin x}{1\,+\,\sin x}$

The left side is: .\(\displaystyle \L\left(\frac{1}{\cos x}\,-\,\frac{\sin x}{\cos x}\right)^2\;=\;\left(\frac{1\,-\,\sin x}{\cos x}\right)^2\;=\;\frac{(1\,-\,\sin x)^2}{\cos^2x}\)

. . . \(\displaystyle \L\frac{(1\,-\,\sin x)(1\,-\,\sin x)}{1\,-\,\sin^2x}\:=\:\frac{(1\,-\,\sin x)(1\,-\,\sin x)}{(1\,-\,\sin x)(1\,+\,\sin x)} \:=\:\frac{1\,-\,\sin x}{1\,+\,\sin x}\)


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