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Problem
If θ is the measure of an acute positive angle in its standard position and determines the Point \( B(\tfrac{3}{5}, y) \) is on the unit circle. Find:
(1) \( \tan (90°-\theta) + \sec (90°-\theta) \)
(2) \( \cot (270° + \theta) - \tan (90° + \theta) - \sin (180° + \theta) \)
Solution
For any point on the unit circle, \( x^2 + y^2 = 1 \):
\( (\tfrac{3}{5})^2 + y^2 = 1 \Rightarrow \tfrac{9}{25} + y^2 = 1 \)
\( y^2 = \tfrac{16}{25} \Rightarrow y = \tfrac{4}{5} \) (since \( y>0 \)).
Point \( B = (\tfrac{3}{5}, \tfrac{4}{5}) \)
Solution of Problem (1)
\( \tan (90°-\theta) + \sec (90°-\theta) \)
= \( \cot \theta + \csc \theta \)
= \( \tfrac{\cos \theta}{\sin \theta} + \tfrac{1}{\sin \theta} \)
= \( \tfrac{3}{4} + \tfrac{5}{4} = \tfrac{8}{4} = 2 \)
Solution of Problem (2)
\( \cot (270° + \theta) - \tan (90° + \theta) - \sin (180° + \theta) \)
= \( -\tan \theta - (-\cot \theta) - (-\sin \theta) \)
= \( -\tfrac{4}{3} + \tfrac{3}{4} + \tfrac{4}{5} \)
= \( \tfrac{-80 + 45 + 48}{60} = \tfrac{13}{60} \)
