# Proof of Basic Trigonometric Identities

In $\displaystyle \vartriangle ABC$,

$\displaystyle \sin \theta =\frac{a}{c}\ \ \ \ \ \ \ \ \cos \theta =\frac{b}{c}\ \ \ \ \ \ \ \ \tan \theta =\frac{a}{b}$

$\displaystyle \cot \theta =\frac{b}{a}\ \ \ \ \ \ \ \ \sec \theta =\frac{c}{b}\ \ \ \ \ \ \ \ \operatorname{cosec}\theta =\frac{c}{a}$

$\displaystyle \tan \theta =\frac{a}{b}=\frac{{\displaystyle \frac{a}{c}}}{{\displaystyle \frac{b}{c}}}=\frac{{\sin \theta }}{{\cos \theta }}$

$\displaystyle \cot \theta =\frac{b}{a}=\frac{{\displaystyle \frac{b}{c}}}{{\displaystyle \frac{a}{c}}}=\frac{{\cos \theta }}{{\sin \theta }}$

$\displaystyle \cot \theta =\frac{b}{a}=\frac{{\displaystyle \frac{b}{b}}}{{\displaystyle \frac{a}{b}}}=\frac{1}{{\tan \theta }}$

$\displaystyle \sec \theta =\frac{c}{b}=\frac{{\displaystyle \frac{c}{c}}}{{\displaystyle \frac{b}{c}}}=\frac{1}{{\cos \theta }}$

$\displaystyle \operatorname{cosec}\theta =\frac{c}{a}=\frac{{\displaystyle \frac{c}{c}}}{{\displaystyle \frac{a}{c}}}=\frac{1}{{\sin \theta }}$

$\displaystyle \begin{array}{l}\ \ \ \text{By Pythagoras'}\ \text{ theorem,}\\\\\ \ \ {{a}^{2}}+{{b}^{2}}={{c}^{2}}\end{array}$

$\displaystyle \therefore \frac{{{{a}^{2}}}}{{{{c}^{2}}}}+\frac{{{{b}^{2}}}}{{{{c}^{2}}}}=1$

$\displaystyle \therefore {{\left( {\frac{a}{c}} \right)}^{2}}+{{\left( {\frac{b}{c}} \right)}^{2}}=1$

$\displaystyle \therefore \ \begin{array}{|l|} \hline \displaystyle {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\ \hline\end{array}$

$\displaystyle \ \ \ \text{Since}\ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$

$\displaystyle \ \ \ \frac{{{{{\sin }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\cos }}^{2}}\theta }}=\frac{1}{{{{{\cos }}^{2}}\theta }}$

$\displaystyle \therefore \ \begin{array}{|l|} \hline \displaystyle {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \\ \hline\end{array}$

$\displaystyle \ \ \ \text{Since}\ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1,$

$\displaystyle \ \ \ \frac{{{{{\sin }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}+\frac{{{{{\cos }}^{2}}\theta }}{{{{{\sin }}^{2}}\theta }}=\frac{1}{{{{{\sin }}^{2}}\theta }}$

$\displaystyle \therefore \ \begin{array}{|l|} \hline \displaystyle 1+{{\cot }^{2}}\theta ={{\operatorname{cosec}}^{2}}\theta \\ \hline\end{array}$