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If cosec q - sin q = a^3 and sec q - cos q = b^3,

prove that (a^2 + b^2) = 1/(ab)^2
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cosec q - sin q = a^3

Multiply both sides by sin(q) to get

1 - sin^2(q) = a^3 sin(q)

Similarly for sec q - cos q = b^3 to get

1 - cos^2(q) = b^3 cos(q)

Using cos^2 + sin^2 = 1 these become

cos^2(q) = a^3 sin(q)
sin^2(q) = b^3 cos(q)

Divide them into one another gives you a sin(q) = b cos(q). Adding them instead gets you 1 = a^3 sin(q)+ b^3 cos(q)

Using a sin(q) = b cos(q) to remove the cos(q) term gives

1 = a^3 sin(q)+ ab^2 sin(q)
1 = asin(q)(a^2 + b^2)

ie (a^2+b^2) = 1/(a sin(q)). I could also have derived (a^2+b^2) = 1/(bcos(q)).

Rather than do all it for you, what would your next step be?
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