Question

Prove the following:
\[\left[ 1+\cot \theta -\sec \left( \dfrac{\pi }{2}+\theta \right) \right]\cdot \left[ 1+\cot \theta +\sec \left( \dfrac{\pi }{2}+\theta \right) \right]=2\cot \theta \]

Answer 1

Hint: We will simplify the left hand side of the given equation. We expect the simplified form of the LHS to be equal to the RHS in the given equation. We will use some trigonometric relations and identities to carry out the simplification. We will also use some algebraic identities to obtain the desired answer.

Complete step by step answer:
We will start with the left hand side that is L.H.S. We know that secant is the reciprocal of cosine. Substituting secant function with the cosine function, we get the following
\[\begin{align}
  & =\left[ 1+\cot \theta -\sec \left( \dfrac{\pi }{2}+\theta \right) \right]\cdot \left[ 1+\cot \theta +\sec \left( \dfrac{\pi }{2}+\theta \right) \right] \\
 & =\left[ 1+\cot \theta -\dfrac{1}{\cos \left( \dfrac{\pi }{2}+\theta \right)} \right]\cdot \left[ 1+\cot \theta +\dfrac{1}{\cos \left( \dfrac{\pi }{2}+\theta \right)} \right] \\
\end{align}\]
Now, we know that there are relations between the angles of trigonometric functions. On simplifying the angles of the trigonometric functions, we get the following result
\[=\left[ 1+\cot \theta +\dfrac{1}{\sin \theta } \right]\cdot \left[ 1+\cot \theta -\dfrac{1}{\sin \theta } \right]\]
Now, using the algebraic identity, we get the following result
\[\begin{align}
  & \left[ (a+b)(a-b)={{a}^{2}}-{{b}^{2}} \right] \\
 & =\left[ {{\left( 1+\cot \theta \right)}^{2}}-{{\left( \dfrac{1}{\sin \theta } \right)}^{2}} \right] \\
 & =\left[ 1+{{\cot }^{2}}\theta +2\cot \theta -{{\left( \dfrac{1}{\sin \theta } \right)}^{2}} \right] \\
 & =\left[ 1+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-\dfrac{1}{{{\sin }^{2}}\theta }+2\cot \theta \right] \\
 & =\left[ \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta -1}{{{\sin }^{2}}\theta } \right]+2\cot \theta \\
\end{align}\]
Now, we use the trigonometric identity that is mentioned, as follows
\[\begin{align}
  & \left[ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right] \\
 & =\left[ \dfrac{1-1}{{{\sin }^{2}}\theta } \right]+2\cot \theta \\
 & =2\cot \theta \\
\end{align}\]
We got the simplified form of the LHS to be equal to the right hand side, that is RHS. Hence, we have proved the given equation.

Note: Simplification of the given expression is the main part of this question. We use various trigonometric and algebraic identities and relations for this simplification. The key aspect is to choose the correct identities for the same. An incorrect choice may lead us into loops and hence, away from the required answer.