Trigonometry! #65

Geometry Level 2

If tan A tan B = x \tan A - \tan B = x cot B cot A = y \cot B - \cot A = y then cot ( A B ) \cot(A-B) is

This problem is part of the set Trigonometry .

1 x 1 y + 1 \frac {1}{x} - \frac {1}{y} + 1 1 x + 1 y \frac {1}{x} + \frac {1}{y} 1 x 1 y 1 \frac {1}{x} - \frac {1}{y} - 1 1 x 1 y \frac {1}{x} - \frac {1}{y}

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2 solutions

Omkar Kulkarni
Feb 21, 2015

cot B cot A = y \cot B-\cot A=y

1 tan B 1 tan A = y \frac{1}{\tan B}-\frac{1}{\tan A}=y

tan A tan B tan A tan B = y \frac{\tan A-\tan B}{\tan A\tan B}=y

x tan A tan B = y \frac{x}{\tan A\tan B}=y

tan A tan B = x y \tan A \tan B=\frac{x}{y}

cot ( A B ) \therefore \cot (A-B)

= 1 tan ( A B ) = \frac{1}{\tan (A-B)}

= 1 + tan A tan B tan A tan B = \frac{1+\tan A\tan B}{\tan A - \tan B}

= 1 + x y x = \frac{1+\frac{x}{y}}{x}

= x + y x y =\frac{x+y}{xy}

= 1 x + 1 y =\boxed{\frac{1}{x}+\frac{1}{y}}

Ramesh Goenka
Jan 31, 2015

break into sin, cos .. use angle sum formulas and manipulate .. !! :D

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