ANGLES IN QUADRANTS

ANGLES IN QUADRANTS
Trigonometric function of complementary Angles
    Consider a right angled triangle ABC, right angled at B

Let  \(\angle A=\theta\) \(\therefore \angle c = 90 - \theta \) from right angle triangle ABC
A)\(\sin \theta = \frac{{AB}}{{AC}}\)
\(\cos \theta = \frac{{AB}}{{AC}}\)
\(\tan \theta = \frac{{BC}}{{AB}}\)
\(\cos ec\theta = \frac{{AC}}{{BC}}\)
\(\sec \theta = \frac{{AC}}{{AB}}\)
\(\cot \theta = \frac{{AB}}{{BC}}\)
B)\(\sin ({90^0} - \theta ) = \frac{{AB}}{{AC}}\)
\(\cos ({90^0} - \theta ) = \frac{{BC}}{{AB}}\)
\(\tan ({90^0} - \theta ) = \frac{{AB}}{{BC}}\)
\(\csc ({90^0} - \theta ) = \frac{{AC}}{{AB}}\)
\(\sec ({90^0} - \theta ) = \frac{{BC}}{{AC}}\)
\(\cot ({90^0} - \theta ) = \frac{{BC}}{{AB}}\)
c)\(\sin ({90^0} - \theta ) = \cos \theta \)
\(\cos ({90^0} - \theta ) = \sin \theta \)
\(\tan ({90^0} - \theta ) = \cot \theta \)
\(\csc ({90^0} - \theta ) = \sec \theta \)
\(\sec ({90^0} - \theta ) = \csc \theta \)
\(\cot ({90^0} - \theta ) = \tan \theta \)