Question

For the given expression prove that LHS is equal to the RHS. The expression is:
$sin\;\theta \left( 1+tan\;\theta \right)+cos\;\theta \;\left( 1+cot\;\theta \; \right)=\;sec\;\theta +\;cosec\;\theta$

Answer 1

Hint: Using the identity $si{{n}^{2}}\theta +co{{s}^{2}}\theta \;=1$, $\dfrac{1}{\cos \theta }=\sec \theta \,\,and\,\,\,\dfrac{1}{\sin \theta }=cosec\;\theta$ we can simplify the LHS and make it equal to the RHS.

Complete step by step answer:
In the given problem, we have to prove that $sin\;\theta \left( 1+tan\;\theta \right)+cos\;\theta \;\left( 1+cot\;\theta \; \right)=\;sec\;\theta +\;cosec\;\theta$
So now we will start with the LHS.
$\Rightarrow sin\;\theta \left( 1+tan\;\theta \right)+cos\;\theta \;\left( 1+cot\;\theta \; \right)$
Now, on simplifying we have:
$\Rightarrow sin\;\theta \left( 1+tan\;\theta \right)+cos\;\theta \;\left( 1+cot\;\theta \; \right)$
Now, it is known that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $cot\;\theta =\dfrac{\cos \theta }{\sin \theta }$, so we can further simplify, as follows:
$\Rightarrow \sin \theta \left( 1+\dfrac{\sin \theta }{\cos \theta } \right)+\cos \theta \left( 1+\dfrac{\cos \theta }{\sin \theta } \right)$
Now, taking the LCM and simplifying the numerator part, we have:

& \Rightarrow \dfrac{\left( \cos \theta +\sin \theta \right){{\cos }^{2}}\theta +\left( \cos \theta +\sin \theta \right){{\sin }^{2}}\theta }{\cos \theta \sin \theta } \\\ & \Rightarrow \dfrac{\left( \cos \theta +\sin \theta \right)({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}{\cos \theta \sin \theta } \\\ & \Rightarrow \dfrac{\left( \cos \theta +\sin \theta \right)(1)}{\cos \theta \sin \theta }\,\,\,\,\,\,\,\,\,\,\,\,\because {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\\ & \Rightarrow \dfrac{1}{\cos \theta }+\dfrac{1}{\sin \theta } \\\ & \Rightarrow \;sec\;\theta +\;cosec\;\theta \,\,\,\,\,\,\,\,\,\,\because \dfrac{1}{\cos \theta }=\sec \theta \,\,and\,\,\,\dfrac{1}{\sin \theta }=cosec\;\theta \\\ & \Rightarrow RHS \\\ \end{aligned}$$ So, here we have finally proved that the LHS is equal to the RHS. Hence, $$sin\;\theta \left( 1+tan\;\theta \right)+cos\;\theta \;\left( 1+cot\;\theta \; \right)=\;sec\;\theta +\;cosec\;\theta $$ is proved. Note: It is important that we don’t get confused between $$\sec \theta $$ and $$\cos \theta $$. Here they are reciprocal of each other. Also, it can be noted that we can have any value of $$\theta $$, as $$\theta $$ value will not affect the identity $$si{{n}^{2}}\theta +co{{s}^{2}}\theta \;=1$$.