Respuesta :

Answer:

See the work below.

Step-by-step explanation:

We can proof this trigonometry equation using the trigonometry identities as follows:

[tex]\boxed{\displaystyle sec\ A=\frac{1}{cos\ A} }[/tex]

[tex]\boxed{\displaystyle tan\ A=\frac{sin\ A}{cos\ A} }[/tex]

[tex]\displaystyle \frac{1}{sec\ A-tan\ A} =\frac{1}{\frac{1}{cos\ A} -\frac{sin\ A}{cos\ A} }[/tex]

                       [tex]\displaystyle=\frac{1}{\frac{1-sin\ A}{cos\ A}}[/tex]

                       [tex]\displaystyle=\frac{cos\ A}{1-sin\ A} \times\frac{1+sin\ A}{1+sin\ A}[/tex]

                       [tex]\displaystyle=\frac{cos\ A(1+sin\ A)}{1-sin^2A}[/tex]

                       [tex]\displaystyle=\frac{cos\ A(1+sin\ A)}{cos^2A}[/tex]

                       [tex]\displaystyle=\frac{1+sin\ A}{cos\ A}[/tex]

                       [tex]\displaystyle=\frac{1}{cos\ A}+\frac{sin\ A}{cos\ A}[/tex]

                       [tex]=sec\ A+tan\ A[/tex]

rvkacademic

Answer:

Proof below

Step-by-step explanation:

We are asked to prove

[tex]\dfrac{1}{\sec A - \tan A} = \sec A + \tan A[/tex]  

  • To prove this, we’ll start by simplifying the left-hand side (LHS):
    [tex]\dfrac{1}{\sec A - \tan A}[/tex]     [1]

  • Use the Pythagorean identity:
    [tex]\:\tan ^2\left(x\right)+1=\sec ^2\left(x\right)[/tex]    [tex]\rightarrow\quad1=\sec ^2\left(x\right)-\tan ^2\left(x\right)[/tex]
  • Substituting for 1 in expression [1] we get
    [tex]\dfrac{\sec ^2A-\tan ^2A}{\sec A-\tan A}[/tex]

  • Apply difference of two squares formula: [tex]x^2-y^2=\left(x+y\right)\left(x-y\right)[/tex][tex]\sec^2 A - \tan^2 A = (\sec A + \tan A)(\sec A - \tan A)[/tex]

  • Replace numerator with the right side of expression above[tex]=\dfrac{(\sec A +\tan A)(\sec A-\tan A)}{\sec A-\tan A}[/tex]

  • Cancel the common factor:  sec(A) -tan(A)
    [tex]= \sec A + \tan A[/tex]

Proved

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