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gogolik [260]
3 years ago
6

Prove the following identity true. Show work.

Mathematics
2 answers:
Elodia [21]3 years ago
8 0

Step-by-step explanation:

Answer is given above

Formulas used

a^2 - b^2 = (a+b)(a-b)

cos^2 t = 1 - sin^2 t

yulyashka [42]3 years ago
3 0

Answer:

The  Pythagorean identity states that

\sin^2 t + \cos^2 t = 1

Using that we can rewrite the left denominator as:

1 - \sin^2 t

Which can be factored as

(1 - \sin t)(1+ \sin t)

The numerator we can expand as:

(1 - \sin t)(1 - \sin t)

On the right hand side, let's multply numerator and denominator with (1 - sin t):

The total formula then becomes:

\dfrac{(1-\sin t)(1-\sin t)}{(1 - \sin t)(1 + \sin t)} = \dfrac{(1-\sin t)(1 - \sin t)}{(1+\sin t)(1 - \sin t)}

There you go... left and right are equal.

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           -------
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or....
1/9    >>>      8/72     (72 is the common denominator)  (Multiply the N and D by 8)
29/8  >>>  261/72     (72 is the common denominator)  (Multiply the N and D by 9)
                 -------------
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