Trigonometric Functions and Identities
4.0 Trigonometric Identities
4.0 Trigonometric Identities
An equation involving trigonometric functions which is true for all those angles for which the functions are defined is called trigonometrical identity. Some identities are -
- $\sin \theta = \frac{1}{{{\text{cosec}}\theta }}$
- $\cos \theta = \frac{1}{{{\text{sec}}\theta }}$
- $\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{1}{{\cot \theta }}$
- ${\sin ^2}\theta + {\cos ^2}\theta = 1$
- ${\tan ^2}\theta + 1 = {\sec ^2}\theta $
Proof: As, we know that $$\tan \theta = \frac{{\sin \theta }}{{\cos \theta }}...(1)$$$$\cos \theta = \frac{1}{{{\text{sec}}\theta }} \Rightarrow \sec \theta = \frac{1}{{\cos \theta }}...(2)$$ From identity $(4)$, we have $${\sin ^2}\theta + {\cos ^2}\theta = 1$$Take ${{{\cos }^2}\theta }$ common, we get $$\begin{equation} \begin{aligned} {\cos ^2}\theta \left( {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} + 1} \right) = 1 \\ {\left( {\frac{{\sin \theta }}{{\cos \theta }}} \right)^2} + 1 = \frac{1}{{{{\cos }^2}\theta }} \\\end{aligned} \end{equation} $$ From $(1)$ and $(2)$, we get $${\tan ^2}\theta + 1 = {\sec ^2}\theta $$
6. $1 + {\cot ^2}\theta = {\text{cose}}{{\text{c}}^2}\theta $
Proof: As, we know that $$\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}...(3)$$$$\sin \theta = \frac{1}{{{\text{cosec}}\theta }} \Rightarrow {\text{cosec}}\theta = \frac{1}{{\sin \theta }}...(4)$$ From identity $(4)$, we have $${\sin ^2}\theta + {\cos ^2}\theta = 1$$Take ${{{\sin }^2}\theta }$ common, we get $$\begin{equation} \begin{aligned} {\sin ^2}\theta \left( {1 + \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right) = 1 \\ 1 + {\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)^2} = \frac{1}{{{{\sin }^2}\theta }} \\\end{aligned} \end{equation} $$ From $(3)$ and $(4)$, we get $$1 + {\cot ^2}\theta = {\text{cose}}{{\text{c}}^2}\theta $$
Question 1. If $\sin \theta = \frac{1}{2}$, and $\theta$ lies in second quadrant, find the value of cos$\theta $, tan$\theta $, cot$\theta $, sec$\theta $, cosec$\theta $.
Solution:
Given : $$\sin \theta = \frac{1}{2}$$
As we know, $$\begin{equation} \begin{aligned} {\cos ^2}\theta = 1 - {\sin ^2}\theta = 1 - \frac{1}{4} \\ {\cos ^2}\theta = \frac{3}{4} \\ \Rightarrow \cos \theta = \pm \frac{{\sqrt 3 }}{2} \\\end{aligned} \end{equation} $$ Since, it is given that $\theta$ lies in second quadrant, means $cos \theta$ is negative. Therefore, we get $$\cos \theta = - \frac{{\sqrt 3 }}{2}$$ Now, $$\tan \theta = \frac{{\sin \theta }}{{\cos \theta }} = \frac{{1/2}}{{ - \sqrt 3 /2}} = - \frac{1}{{\sqrt 3 }}$$ Since, $\theta$ lies in second quadrant, $tan \theta$ is negative. Now, $$\cot \theta = \frac{1}{{\tan \theta }} = \frac{1}{{\frac{{ - 1}}{{\sqrt 3 }}}} = - \sqrt 3 $$$${\text{cosec}}\theta = \frac{1}{{\sin \theta }} = \frac{1}{{\frac{1}{2}}} = 2$$$${\text{sec}}\theta = \frac{1}{{\cos \theta }} = \frac{1}{{ - \frac{{\sqrt 3 }}{2}}} = - \frac{2}{{\sqrt 3 }}$$
