TRIGONOMETRIC RATIOS
Trigonometric function of complementary angles
1. Consider a right angled triangle ABC, right angled at B. Let \(\angle A = \theta \) then \(\angle C = \left( {90 - \theta } \right)\)From the diagram.
\(\sin \theta = \frac{{BC}}{{AC}}\)
\(\sin \left( {90 - \theta } \right) = \frac{{AB}}{{AC}}\)
\(\cos \theta = \frac{{AB}}{{AC}}\)
\(\cos \left( {90 - \theta } \right) = \frac{{BC}}{{AC}}\)
\(\tan \theta = \frac{{BC}}{{AB}}\)
\(\tan \left( {90 - \theta } \right) = \frac{{AB}}{{BC}}\)
\(\cos ec\theta = \frac{{AC}}{{BC}}\)
\(\cos ec\left( {90 - \theta } \right) = \frac{{AC}}{{AB}}\)
\(\sec \theta = \frac{{AC}}{{AB}}\)
\(\sec \left( {90 - \theta } \right) = \frac{{AC}}{{BC}}\)
\(\cot \theta = \frac{{AB}}{{BC}}\)
\(\cot \left( {90 - \theta } \right) = \frac{{BC}}{{AB}}\)
From the above results,
\(\begin{gathered}
\sin \left( {90 - \theta } \right) = \cos \theta , \hfill \\
\cos ec\left( {90 - \theta } \right) = \sec \theta \hfill \\
\cos \left( {90 - \theta } \right) = \sin \theta , \hfill \\
\sec \left( {90 - \theta } \right) = \cos ec\theta \hfill \\
\tan \left( {90 - \theta } \right) = \cot \theta , \hfill \\
\cot \left( {90 - \theta } \right) = \tan \theta \hfill \\
\end{gathered} \)
2. Sign of Trigonometrci Ratios.
If ‘\(\theta\)’ lies in quadrants \({Q_1},{Q_2},{Q_3},{Q_4}\), then the sign of trigonometric ratios are as follows:
Note :
1. \({0^0},{90^0},{180^0},{270^0},{360^0},.........\) are called quadrant angles.
2. With the phrase All Silver Tea Cups we can remember the sign of trigometric ratios.
Note :
1. For \({0^0},{0^0} \pm \theta ,{180^0} \pm \theta ,{360^0} \pm \theta ,\) there is no change in the trigonometric ratios.
2.\({90^0} \pm \theta ,{270^0} \pm \theta ,\) the change in the trigonometric ratios is as follows:
\(\sin \leftrightarrow \cos ,\tan \leftrightarrow \cot ,\sec \leftrightarrow \cos ec\)
3. Whether we get + or - sign in the answer, it should be decided by considering the quadrant in which the angle \(\left( {n{{.360}^0} \pm \theta } \right)\) lies.