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16th July 2008 - 10:12 AM
If cosec q - sin q = a^3 and sec q - cos q = b^3,
prove that (a^2 + b^2) = 1/(ab)^2
16th July 2008 - 01:28 PM
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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