How can I confidently find the value of z in the given equation?

[SweatingBear]

Find the value of \(\displaystyle z = \frac 1{\cos x} + \tan x\) given \(\displaystyle \frac 1{\cos x} - \tan x = 2\).

By squaring and using trigonometric identities on the given equation, one can arrive at

\(\displaystyle \frac {1- \sin x}{1 + \sin x} = 4\)

The unknown expression \(\displaystyle z\) can, by squaring, be written as

\(\displaystyle z^2 = \frac {1 + \sin x}{1 - \sin x} = \frac 14 \)

and thus

\(\displaystyle z = \pm \frac 12\)

However I am unable to rule out the incorrect value for \(\displaystyle z\); how can I from here confidently find the actual correct value for \(\displaystyle z\)?

Another way to solve this is to multiply \(\displaystyle z\) with \(\displaystyle \frac 1{\cos x} - \tan x\) and expand to obtain \(\displaystyle z = \frac 12\). Sure it works and is unambiguous, but I persistently want to be able to obtain the same answer through my previous approach (and thus get better at analyzing the underpinning algebra).

Forum help me rule out the solution \(\displaystyle z = - \frac 12\) from my previous solution.

 

[MarkFL]

One way you could eliminate the extraneous solution is to use:

\(\displaystyle z=-\frac{1}{2}=\frac{\cos(x)}{\sin(x)-1}\)

in the first equation as a check:

\(\displaystyle \frac{\cos(x)}{\sin(x)-1}=\frac{\sin(x)+1}{\cos(x)}\)

This gives us:

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

and since this is not an identity, we must discard \(\displaystyle z=-\frac{1}{2}\) as a solution.

 

[Opalg]

Find the value of \(\displaystyle z = \frac 1{\cos x} + \tan x\) given \(\displaystyle \frac 1{\cos x} - \tan x = 2\).

By squaring and using trigonometric identities on the given equation, one can arrive at

\(\displaystyle \frac {1- \sin x}{1 + \sin x} = 4\)

Your equation \(\displaystyle \frac {1- \sin x}{1 + \sin x} = 4\) tells you that $1 - \sin x = 4(1 + \sin x)$, from which $\sin x = -3/5.$ Now write the equation \(\displaystyle \frac 1{\cos x} - \tan x = 2\) as \(\displaystyle \frac {1 - \sin x}{\cos x} = 2\) to deduce that $\cos x = 4/5$. Then you can calculate \(\displaystyle \frac 1{\cos x} + \tan x = \frac {1 + \sin x}{\cos x} = \frac12.\)

 

[SweatingBear]

One way you could eliminate the extraneous solution is to use:

\(\displaystyle z=-\frac{1}{2}=\frac{\cos(x)}{\sin(x)-1}\)

I am having a sudden brain freeze: Where did that equality come from? We had \(\displaystyle z = \frac {1 + \sin x}{\cos x}\) and not \(\displaystyle z = \frac{\cos(x)}{\sin(x)-1}\), or am I just barking up the wrong tree?

Your equation \(\displaystyle \frac {1- \sin x}{1 + \sin x} = 4\) tells you that $1 - \sin x = 4(1 + \sin x)$, from which $\sin x = -3/5.$ Now write the equation \(\displaystyle \frac 1{\cos x} - \tan x = 2\) as \(\displaystyle \frac {1 - \sin x}{\cos x} = 2\) to deduce that $\cos x = 4/5$. Then you can calculate \(\displaystyle \frac 1{\cos x} + \tan x = \frac {1 + \sin x}{\cos x} = \frac12.\)

All right thanks a lot, but one question: How come the act of squaring the equation did, this time, not produce any extraneous information? I would, understandably so, actually expect that to occur.

 

[MarkFL]

I am having a sudden brain freeze: Where did that equality come from? We had \(\displaystyle z = \frac {1 + \sin x}{\cos x}\) and not \(\displaystyle z = \frac{\cos(x)}{\sin(x)-1}\), or am I just barking up the wrong tree?...

We are originally given:

\(\displaystyle \frac{1}{\cos(x)}-\tan(x)=2\)

\(\displaystyle \frac{1-\sin(x)}{\cos(x)}=2\)

Negating and inverting, we may write:

\(\displaystyle \frac{\cos(x)}{\sin(x)-1}=-\frac{1}{2}\)

Now, we may use this value for \(\displaystyle z=-\frac{1}{2}\) to see if the equation results in an identity or not.

 

[SweatingBear]

Thank you MarkFL! That showed me how to carefully check the algebra.

@Opalg: Would really appreciate if you could share your thoughts on:

Your equation \(\displaystyle \frac {1- \sin x}{1 + \sin x} = 4\) tells you that $1 - \sin x = 4(1 + \sin x)$, from which $\sin x = -3/5.$ Now write the equation \(\displaystyle \frac 1{\cos x} - \tan x = 2\) as \(\displaystyle \frac {1 - \sin x}{\cos x} = 2\) to deduce that $\cos x = 4/5$. Then you can calculate \(\displaystyle \frac 1{\cos x} + \tan x = \frac {1 + \sin x}{\cos x} = \frac12.\)

All right thanks a lot, but one question: How come the act of squaring the equation did, this time, not produce any extraneous information? I would, understandably so, actually expect that to occur.

 

[Opalg]

How come the act of squaring the equation did, this time, not produce any extraneous information? I would, understandably so, actually expect that to occur.

It's not entirely clear to me why that is. But notice that the original equation that you squared was \(\displaystyle \frac 1{\cos x} - \tan x = 2\). Writing this as \(\displaystyle \frac{1-\sin x}{\cos x}\), you see that the numerator must be positive (because the sin function cannot be bigger than $1$). Since $1-\sin x$ cannot be negative, its square has only one square root available (namely the positive one).

That does not completely explain why no extraneous solutions occur, but I think it must be at least a partial reason.

 

[SweatingBear]

It's not entirely clear to me why that is. But notice that the original equation that you squared was \(\displaystyle \frac 1{\cos x} - \tan x = 2\). Writing this as \(\displaystyle \frac{1-\sin x}{\cos x}\), you see that the numerator must be positive (because the sin function cannot be bigger than $1$). Since $1-\sin x$ cannot be negative, its square has only one square root available (namely the positive one).

That does not completely explain why no extraneous solutions occur, but I think it must be at least a partial reason.

Thanks for the reply. That's a good analysis and to build upon it, I would actually want to claim \(\displaystyle \cos x > 0\) because

\(\displaystyle \frac {1-\sin x}{\cos x} = 2 > 0 \implies \frac {1- \sin x }{\cos x} > 0\)

and since \(\displaystyle 1 -\sin x > 0\) the quotient will be positive if and only if \(\displaystyle \cos x > 0\). So we are only getting the positive square root for the whole quotient and hence do not have to worry about potential extraneous roots. Right?