Trigonometric Identities

Go back to  'Trigonometry'

Introduction:

An equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angles involved.
There are three improtant trigonometric identities which are extensively used throughout the topic of trigonometry. These are as follows:

  •    \({\sin ^2}\theta + {\cos ^2}\theta = 1\)
  •   \(1 + {\tan ^2}\theta = {\sec ^2}\theta \)
  •    \(1 + {\cot ^2}\theta = co{\sec ^2}\theta \)

✍Note: Please go through basic properties of trigonometric ratios to understand the proof of above identities.
Now, Let's solve some basic problems based on these three trigonometric identities.


Solved Examples:

Example 1: Prove the following identity :

\[\frac{{{{\sin }^3}\theta + {{\cos }^3}\theta }}{{\sin \theta + \cos \theta}}+ \sin \theta\; \cos\theta = 1\]

Solution: We make use of the following identity:

\[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\]

For the given identity, we have

\[\begin{align}& LHS = \left\{ {\frac{{\left( {\sin \theta + \cos \theta } \right)\left( {{{\sin }^2}\theta - \sin \theta cos\theta + co{s^2}\theta } \right)}}{{\sin \theta + \cos \theta + \sin \theta\; cos\theta}}} \right\}\\ &\qquad\,= ({\sin ^2}\theta - \sin \theta \cos \theta + {\cos ^2}\theta )\,\, + \sin \theta \cos \theta \\ &\qquad\,= {\sin ^2}\theta + {\cos ^2}\theta=\;1\;= RHS \end{align}\]

Hence Proved.


Example 2: Prove that :

\[{(\sin \theta + cosec\; \theta )^2} + {(\cos \theta + \sec \theta )^2} = 7 + {\tan ^2}\theta + {\cot ^2}\theta \]

Solution: We have

\[\begin{align}
  LHS &= {\sin ^2}\theta  + {\text{cosec}^2}\theta  + \underbrace {2\sin \theta \text{cosec}\;\theta }_{ = 1} \hfill \\
   &\,\,\,\,\,\,+ {\cos ^2}\theta  + {\sec ^2}\theta  + \underbrace {2\cos \theta \sec \theta }_{ = 1} \hfill \\
   &= ({\sin ^2}\theta  + {\cos ^2}\theta ) + (1 + {\cot ^2}\theta ) \hfill \\
   &\,\,\,\,\,\,+ (1 + {\tan ^2}\theta ) + 2 + 2 \hfill \\
   &= 7 + {\tan ^2}\theta  + {\cot ^2}\theta \; = RHS \hfill \\ 
\end{align} \]

Hence Proved.


Example 3: Prove that:

\[\left( {1 + \frac{1}{{{{\tan }^2}\theta }}} \right)\left( {1 + \frac{1}{{{{\cot }^2}\theta }}} \right) = \frac{1}{{{{\sin }^2}\theta - {{\sin }^4}\theta }}\]

Solution: The LHS is

\[\begin{align}&LHS = (1 + {\cot ^2}\theta )\left( {1 + {{\tan }^2}\theta } \right)\\ &\qquad\;= {\text{cosec}^2}\theta \times {\sec ^2}\theta \\ &\qquad\;= \frac{1}{{{{\sin }^2}\theta\, {{\cos }^2}\theta }}\\ &\qquad\;= \frac{1}{{{{\sin }^2}\theta\; (1 - {{\sin }^2}\theta )}}\\ &\qquad\;= \frac{1}{{{{\sin }^2}\theta - {{\sin }^4}\theta }} \;= RHS\end{align}\]

Hence Proved.


Example 4: Prove that:

\[\frac{{1 - \cos \theta }}{{1 + \cos \theta }} = {\left( {\text{cosec}\,\theta - \cot \theta } \right)^2}\]

Solution: Let’s start with the right side

\[\begin{align}&RHS = {\left( {\frac{1}{{\sin \theta }} - \frac{{\cos \theta }}{{\sin \theta }}} \right)^2}\\ &\qquad\;\;= {\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)^2}\\ &\qquad\;\;= \frac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{{\sin }^2}\theta }}\\ &\qquad\;\;= \frac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{1 - {{\cos }^2}\theta }}\\ &\qquad\;\;= \frac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{(1 + \cos \theta )(1 - \cos \theta )}}\\ &\qquad\;\;= \frac{{1 - \cos \theta }}{{1 + \cos \theta }} = LHS\end{align}\]

Alternatively, if we have to start with the left side, we can proceed as follows:

\[\begin{align}&LHS = \frac{{1 - \cos \theta }}{{1 + \cos \theta }} \times \frac{{1 - \cos \theta }}{{1 - \cos \theta }}\\ &\qquad\;= \frac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{1 - \cos^2 \theta }} = \frac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{{\sin }^2}\theta }}\\ &\qquad\;= {\left( {\frac{{1 - \cos \theta }}{{\sin \theta }}} \right)^2} = {\left( {\frac{1}{{\sin \theta }} - \frac{{\cos \theta }}{{\sin \theta }}} \right)^2}\\ &\qquad\;= {(\text{cosec}\,\theta - \cot \theta )^2} = RHS\end{align}\]

Both of these approaches are equivalent.
Hence Proved.


Example 5: Prove the following identity:

\[\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \frac{1}{{\sec \theta - \tan \theta }}\]

Solution: We manipulate the left side to express it in terms of \(\sec \theta \) and \(\tan \theta \). We divide both the numerator and denominator on the left side by \(\cos \theta \) :

\[\begin{align}&LHS = \frac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}\\ &\qquad\;= \frac{{\tan \theta + \sec \theta - 1}}{{\tan \theta - \sec\theta + 1}}\end{align}\]

Now we replace the 1 in the numerator with the term \(\left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)\). This will enable us to factorize the numerator:

\[\begin{align}&LHS = \frac{{\left( {\tan \theta + \sec \theta } \right) - \left( {{{\sec }^2}\theta - {{\tan }^2}\theta } \right)}}{{\tan \theta - \sec\theta + 1}}\\ &\qquad\;= \frac{{\left( {\tan \theta + \sec \theta } \right) - \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right)}}{{\tan \theta - \sec \theta + 1}}\\ &\qquad\;= \frac{{\left( {\tan \theta + \sec \theta } \right)\left( {1 - \sec \theta + \tan \theta } \right)}}{{\tan \theta - \sec\theta + 1}}\\ &\qquad\;= \sec \theta + \tan \theta \end{align}\]

Finally we do the following to reach the RHS:

\[\begin{align}&LHS = \sec \theta + \tan \theta = \frac{{\sec \theta + \tan \theta }}{{{{\sec }^2}\theta - {{\tan }^2}\theta }}\\ &\qquad\;= \frac{{\sec \theta + \tan \theta }}{{\left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right)}}\\ &\qquad\;= \frac{1}{{\sec \theta - tan\theta }} = RHS\end{align}\]

Hence Proved.


yesChallenge 1: Prove the following identity:

\[\frac{{\tan \theta }}{{1 - \cot \theta }} + \frac{{\cot \theta }}{{1 - \tan \theta }} = 1 + \sec \theta {\text{ cosec}}\theta \]

⚡Tip: Write the expression in terms of \({\sin \theta }\) and \({\cos \theta }\)

yesChallenge 2: Prove the following identity:

\[\frac{{1 + \sec \theta }}{{\sec \theta }} = \frac{{{{\sin }^2}\theta }}{{1 - \cos \theta }}\]

⚡Tip: Simplify LHS and RHS separately.


Important Topics

Download Trigonometry Worksheets
Trigonometry
Grade 10 | Answers Set 1
Trigonometry
Grade 9 | Questions Set 1
Trigonometry
Grade 9 | Answers Set 1
Trigonometry
Grade 10 | Questions Set 1