Question

Prove that: $\left( \cos ec\theta -\sec \theta \right)\left( \cot \theta -\tan \theta \right)=\left( \cos ec\theta +\sec \theta \right)\left( \cos ec\theta \sec \theta -2 \right)$.

Answer 1

Hint: Change: $\cos ec\theta \text{ into }\dfrac{1}{\sin \theta }$, $\sec \theta \text{ into }\dfrac{1}{\cos \theta }$, \[\cot \theta \text{ into }\dfrac{\cos \theta }{\sin \theta }\] and $\tan \theta \text{ into }\dfrac{\sin \theta }{\cos \theta }$. Then simplify the terms of the left hand side to get the terms of the right hand side or simplify the terms of the right hand side to get the terms of the left hand side.

Complete step-by-step answer:

In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. They are distinct from triangle identities, which are identities involving angles and side lengths of a triangle.
Now, let us come to the question. we have been given,
$L.H.S=\left( \cos ec\theta -\sec \theta \right)\left( \cot \theta -\tan \theta \right)$
Changing, $\cos ec\theta \text{ into }\dfrac{1}{\sin \theta }$, $\sec \theta \text{ into }\dfrac{1}{\cos \theta }$, \[\cot \theta \text{ into }\dfrac{\cos \theta }{\sin \theta }\] and $\tan \theta \text{ into }\dfrac{\sin \theta }{\cos \theta }$, we get,
$L.H.S=\left( \dfrac{1}{\sin \theta }-\dfrac{1}{\cos \theta } \right)\left( \dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\operatorname{co}\operatorname{s}\theta } \right)$
 Now taking L.C.M, we get,
$L.H.S=\left( \dfrac{\cos \theta -\sin \theta }{\sin \theta \cos \theta } \right)\left( \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\sin \theta \cos \theta } \right)$
We know that, ${{a}^{2}}-{{b}^{2}}=(a-b)(a+b)$. Therefore,
$\begin{align}
  & L.H.S=\left( \dfrac{\cos \theta -\sin \theta }{\sin \theta \cos \theta } \right)\left( \dfrac{(\cos \theta -\sin \theta )\left( \cos \theta +\sin \theta \right)}{\sin \theta \cos \theta } \right) \\
 & \text{ =}\dfrac{{{\left( \cos \theta -\sin \theta \right)}^{2}}}{\sin \theta \cos \theta }\times \dfrac{\left( \cos \theta +\sin \theta \right)}{\sin \theta \cos \theta } \\
\end{align}$
Expanding the square term, we get,
$L.H.S=\left( \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\sin \theta \cos \theta }{\sin \theta \cos \theta } \right)\left( \dfrac{\cos \theta +\sin \theta }{\sin \theta \cos \theta } \right)$
We know that, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$.
$\begin{align}
  & \therefore L.H.S=\left( \dfrac{1-2\sin \theta \cos \theta }{\sin \theta \cos \theta } \right)\left( \dfrac{\cos \theta +\sin \theta }{\sin \theta \cos \theta } \right) \\
 & \text{ }=\left( \dfrac{1}{\sin \theta \cos \theta }-\dfrac{2\sin \theta \cos \theta }{\sin \theta \cos \theta } \right)\left( \dfrac{\cos \theta }{\sin \theta \cos \theta }+\dfrac{\sin \theta }{\sin \theta \cos \theta } \right) \\
 & \text{ =}\left( \dfrac{1}{\sin \theta }\times \dfrac{1}{\cos \theta }-\dfrac{2\sin \theta \cos \theta }{\sin \theta \cos \theta } \right)\left( \dfrac{1}{\sin \theta }+\dfrac{1}{\cos \theta } \right) \\
 & \text{ }=\left( \cos ec \theta \sec\theta -2 \right)\left( \cos ec \theta +\ sec\theta \right) \\
 & \text{ }=R.H.S \\
\end{align}$

Note: We can also apply a reverse process to solve this question. That means we can start with the right hand side to simplify the equation and reach towards the left hand side. There can be a third process also. What we can do is, we start solving with L.H.S and come to a certain step and stop. Then, we start solving the R.H.S side and try to get to that step where we had stopped the L.H.S.