25, 35, 55, 65

Geometry Level 2

csc ( 6 5 + θ ) sec ( 2 5 θ ) tan ( 5 5 θ ) + cot ( 3 5 + θ ) = ? \csc( 65^\circ + \theta) - \sec (25^\circ - \theta) - \tan (55^\circ - \theta) + \cot (35^\circ + \theta) = \ ?

This problem is part of the set Trigonometry .

0 0 1 -1 1 1 2 2

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csc ( 6 5 + θ ) sec ( 2 5 θ ) tan ( 5 5 θ ) + cot ( 3 5 + θ ) \csc{(65^\circ+\theta)} - \sec{(25^\circ-\theta)} - \tan{(55^\circ-\theta)} + \cot{(35^\circ+\theta)}

= 1 sin ( 6 5 + θ ) 1 cos ( 2 5 θ ) sin ( 5 5 θ ) cos ( 5 5 θ ) + cos ( 3 5 + θ ) sin ( 3 5 + θ ) = \dfrac {1}{\sin{(65^\circ+\theta)} } - \dfrac {1}{\cos{(25^\circ-\theta)}} - \dfrac {\sin{(55^\circ-\theta)} }{\cos{(55^\circ-\theta)} } + \dfrac {\cos{(35^\circ+\theta)} }{\sin{(35^\circ+\theta)} }

= 1 sin ( 6 5 + θ ) 1 sin ( 6 5 + θ ) sin ( 5 5 θ ) cos ( 5 5 θ ) + sin ( 5 5 θ ) cos ( 5 5 θ ) = \dfrac {1}{\sin{(65^\circ+\theta)} } - \dfrac {1}{\sin{(65^\circ+\theta)} } - \dfrac {\sin{(55^\circ-\theta)} }{\cos{(55^\circ-\theta)} } + \dfrac {\sin{(55^\circ-\theta)} }{\cos{(55^\circ-\theta)} }

= 0 =\boxed{0}

This is how i solved. Take A = (25 - Θ) and B = (35 + Θ) So, the given equation becomes cosec(90-A) - secA -tan(90-A) + cotA = secA - secA - cotA + cotA = 0. :) (My fon't isnt that perfect)!

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