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13 October, 17:17

I) sin^2 A sec^2 B + tan^2 B cos^2 A = sin^2A + tan²B

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  1. 13 October, 17:37
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    sin² A sec² B + tan² B cos² A

    A good first step is to write everything in terms of sine and cosine.

    sin² A / cos² B + sin² B cos² A / cos² B

    The fractions have the same denominator, so combine into one:

    (sin² A + sin² B cos² A) / cos² B

    Using Pythagorean identity, we can rewrite sin² B as 1 - cos² B:

    (sin² A + (1 - cos² B) cos² A) / cos² B

    Distribute:

    (sin² A + cos² A - cos² B cos² A) / cos² B

    Pythagorean identity:

    (1 - cos² B cos² A) / cos² B

    Now divide into two fractions again:

    1 / cos² B - cos² B cos² A / cos² B

    Simplify:

    sec² B - cos² A

    Using Pythagorean identity again:

    (tan² B + 1) - (1 - sin² A)

    tan² B + 1 - 1 + sin² A

    tan² B + sin² A
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