Respuesta :

tramserran

Answer:  See below

Step-by-step explanation:

[tex]\dfrac{1-\sec x}{1+\sec x}\quad =\quad \dfrac{\cos x-1}{\cos x+1}[/tex]

LHS --> RHS

Convert sec x to 1/(cos x):

[tex]\dfrac{1-\dfrac{1}{\cos x}}{1+\dfrac{1}{\cos x}}[/tex]

Find a common denominator for the numerator and denominator:

[tex]\dfrac{\dfrac{\cos x}{\cos x}-\dfrac{1}{\cos x}}{\dfrac{\cos x}{\cos x}+\dfrac{1}{\cos x}}\quad = \quad \dfrac{\dfrac{\cos x-1}{\cos x}}{\dfrac{\cos x+1}{\cos x}}[/tex]

Eliminate the denominators:

[tex]\quad \dfrac{\cos x-1}{\cos x+1}[/tex]

LHS = RHS so identity is proven.

Answer:

[tex]\frac{1-secx}{1+secx}[/tex]    =   [tex]\frac{cosx - 1}{cosx + 1}[/tex]

Step-by-step explanation:

We are to prove that 1 - sec x/1 + sec x is equal to cos x - 1/cos x + 1

To do this, we need to know that secx= 1/cosx

from the question given  (1-secx) ÷ (1+secx)

Lets first simplify (1-secx)

1-secx = 1 - 1/cosx  =  [tex]\frac{cosx - 1}{cosx}[/tex]

Also we will go ahead and simplify (1+secx)

1 + secx = 1 + 1/cosx =[tex]\frac{cosx + 1}{cosx}[/tex]

[tex]\frac{1-secx}{1+secx}[/tex]    =   [tex]\frac{cosx - 1}{cosx}[/tex]  ÷  [tex]\frac{cosx + 1}{cosx}[/tex]

             =  [tex]\frac{cosx - 1}{cosx}[/tex]  × [tex]\frac{cosx}{cosx + 1}[/tex]

(The cosx at the numerator will cancel-out the cosx  at the denominator)

     

             = [tex]\frac{cosx - 1}{cosx + 1}[/tex]

[tex]\frac{1-secx}{1+secx}[/tex]    =   [tex]\frac{cosx - 1}{cosx + 1}[/tex]

Hence proved.

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