Help me with this trigonometric identity

[Princess9000]

How can i solve this??? cos(x)(Sec(x)-1) goes under tan(x)
sec (x)+1= tan (x)
sin (x) cos(x)(Sec(x)-1)

Please help me!!

 

[tkhunny]

What have you tried. There are at least a half dozen things that come immediately to mind that might do something. Let's see what comes to mind for you.

 

[soroban]

Hello, Princess9000!

Please re-state the problem accurately.
As given, it makes no sense.

How can i solve this?
sec (x)+1= tan (x) sen (x) cos(x)(Sec(x)-1)

Please help me!

In the title, you mentioned an identity.
But the statement you gave is not an identity.

Then you want to solve the equation.
This is possible, but what is the equation?

Why are two parts underlined?
Were you trying to indicate some fractions?
If so, exactly what are they?

 

[Deleted member 4993]

How can i solve this???
sec (x)+1= tan (x) sen (x) cos(x)(Sec(x)-1)

Please help me!!

What is sen (x)?

 

[Princess9000]

What have you tried. There are at least a half dozen things that come immediately to mind that might do something. Let's see what comes to mind for you.

Im not sure what to do first. Please help me. The only thing i know i have to times by the inverse! help me!

 

[Princess9000]

What have you tried. There are at least a half dozen things that come immediately to mind that might do something. Let's see what comes to mind for you.

Hello, Princess9000!

Please re-state the problem accurately.
As given, it makes no sense.


In the title, you mentioned an identity.
But the statement you gave is not an identity.

Then you want to solve the equation.
This is possible, but what is the equation?

Why are two parts underlined?
Were you trying to indicate some fractions?
If so, exactly what are they?

They are fractions! It is an identity! please help me

 

[Bob Brown MSEE]

Im not sure what to do first. Please help me. The only thing i know i have to times by the inverse! help me!

It often helps to replace all trig functions with sin and cos equivalent forms.

 

[JeffM]

They are fractions! It is an identity! please help me

Unfortunately, it is not clear what your fractions are. If you do not know LaTeX (and there is no reason that you should), write out fractions using the forward slash or / and parentheses according to PEDMAS. Otherwise we are guessing at what you are asking.

Is this what you meant to show

$\displaystyle \dfrac{sec(x) + 1}{sin(x)} = \dfrac{tan(x)}{cos(x) * [sec(x) - 1]}.$

If so, show that as [sec(x) + 1] / sin(x) = tan(x) / {cos(x) * [sec(x) - 1]}.

Furthermore, what EXACTLY does the problem say that you are to do?

 

[soroban]

Hello, Princess9000!

We are wasting a LOT of time . . .

$\displaystyle \text{Prove: }\:\dfrac{\sec x + 1}{\sin x\cos x} \:=\:\dfrac{\tan x}{1-\sec x}$

This is not an identity! . Try $\displaystyle x = \tfrac{\pi}{4}$


It is not even a solvable equation.

We'll assume that the solutions are on the interval $\displaystyle [0,\,2\pi]$
Note that: .$\displaystyle x \;\ne\;0,\:\tfrac{\pi}{2},\:\pi,\:\tfrac{3\pi}{2},\:2\pi$


Cross-multiply: .$\displaystyle (1 + \sec x)(1 - \sec x) \:=\:\sin x\cos x\tan x$

. . . . . . . . . . . . . . . . . . . . $\displaystyle 1 - \sec^2\!x \:=\:\sin x\cos x\dfrac{\sin x}{\cos x}$

. . . . . . . . . . . . . . . . . . . . . $\displaystyle -\tan^2\!x \:=\:\sin^2\!x$

. . . . . . . . . . . . . . . . .$\displaystyle \sin^2\!x + \tan^2\!x \:=\:0$

We have: the sum of two positive quantities equals zero.
. . This is true only if both quantities are zero.

Hence: .$\displaystyle \sin^2\!x \,=\,0\,\text{ and }\,\tan^2\!x \,=\,0$

But this gives us: .$\displaystyle x \:=\:0,\,\pi,\,2\pi$ .which are disallowed.

Therefore, the equation has no real solutions.