The minimum value of ( tan A − cos B ) 2 + ( cot A − sin B ) 2 , where A and B are independent variables, is of the form a − b c , where a , b , and c are natural numbers and c is square-free. Find the value of a + b + c .
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The given expression can be interpreted as the square of the distance between the points ( tan A , cot A ) and ( cos B , sin B ) . Now the minimum value of this distance is the minimum distance between the curves x y = 1 and x 2 + y 2 = 1 which comes out to be 3 − 2 2
@Rajyawardhan Singh , as tan, cot, cos, and sin are functions, you should put a backslash in front and a space behind \tan A tan A , \cot B cot B , \sin \frac \pi 3 sin 3 π , \cos frac \theta {12} cos 1 2 θ , just like \sum {k=1}^\infty ∑ k = 1 ∞ , \int 0^\frac \pi 2 ∫ 0 2 π . Note that function name is not in italic and there is a space between the function name and argument.
Thank you sir I corrected it
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By Titu's lemma , we have:
( tan A − cos B ) 2 + ( cot A − sin B ) 2 ≥ 2 ( tan A + cot A − cos B − sin B ) 2 ≥ 2 ( 2 tan A cot A − 2 sin ( B + 4 5 ∘ ) ) 2 ≥ 2 ( 2 − 2 ) 2 = 3 − 2 2 By AM-GM inequality: Equality occurs when tan A = cot A Maximum occurs when sin ( B + 4 5 ∘ ) = 1
Therefore a + b + c = 3 + 2 + 2 = 7 .
Reference: AM-GM inequality