# TrigonometryValue of complex expression

#### sweatingbear

##### Member
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

Staff member
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

##### MHB Oldtimer
Staff member
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

##### Member
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

Staff member
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

##### Member
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

##### MHB Oldtimer
Staff member
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

##### Member
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?