Respuesta :

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Answer:

(identity has been verified)

Step-by-step explanation:

Verify the following identity:

(sec(A) + tan(A)) (1 - sin(A)) = cos(A)

Write secant as 1/cosine and tangent as sine/cosine:

(1 - sin(A)) (1/cos(A) + sin(A)/cos(A)) = ^?cos(A)

Put 1/cos(A) + sin(A)/cos(A) over the common denominator cos(A): 1/cos(A) + sin(A)/cos(A) = (sin(A) + 1)/cos(A):

(sin(A) + 1)/cos(A) (1 - sin(A)) = ^?cos(A)

Multiply both sides by cos(A):

(1 - sin(A)) (sin(A) + 1) = ^?cos(A)^2

(1 - sin(A)) (sin(A) + 1) = 1 - sin(A)^2:

1 - sin(A)^2 = ^?cos(A)^2

cos(A)^2 = 1 - sin(A)^2:

1 - sin(A)^2 = ^?1 - sin(A)^2

The left hand side and right hand side are identical:

Answer: (identity has been verified)

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           Hi my lil bunny!

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Let's do this step by step.

Prove that sec A ( 1 - sin A) ( sec A + tan A) = 1

Solving L.H.S

sec A ( 1 - sin A) ( sec A + tan A)

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

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

We know that [tex]( a - b) ( a + b) = a^2 - b^2[/tex]

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

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

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

[tex]= 1[/tex]

[tex]= R . H. S[/tex]

Thus, L.H.S = R.H.S

Hence proved.

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Hope this helped you.

Could you maybe give brainliest..?

❀*May*❀

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