Question

Prove that: $\left( \sin \theta -\cos ec\theta \right)\left( \cos \theta -\sec \theta \right)=\dfrac{1}{\tan \theta +\cot \theta }$

Answer 1

Hint: Simplify the LHS and RHs of the given equation. To simplify LHS, write $\left( \sin \theta -\cos ec\theta \right)\times \left( \cos \theta -\sec \theta \right)$ as $\left( \cos ec\theta -\sin \theta \right)\times \left( \sec \theta -\cos \theta \right)$. Then convert $\cos ec\theta $ and $\sec \theta $ into terms of $\sin \theta $ and $\cos \theta $ respectively, using,

$\cos ec\theta =\dfrac{1}{\sin \theta }and\,\sec \theta =\dfrac{1}{\cos \theta }$, After that use-

${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and then simplify the LHS.

Complete step by step answer:
To simplify RHS, write

$\tan \theta =\dfrac{\sin \theta }{\cos \theta }\,and\,\,\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

And simplify the expression given in RHS. Using ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

We have to just simplify both side of the question.

Now taking LHS= $\left( \sin \theta -\cos ec\theta \right)\left( \cos \theta -\sec \theta \right)$.

Taking (-1) common from both sides, we get-

$\begin{align}

  & LHS=\left( -1 \right)\left( \cos ec\theta -\sin \theta \right)\times \left( -1 \right)\left( \sec \theta -\cos \theta \right) \\

 & LHS=\left( \cos ec\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left[ \because \left( -1 \right)\left( -1 \right)=1 \right] \\

\end{align}$

Using $\cos ec\theta =\dfrac{1}{\sin \theta }$ and $\sec \theta =\dfrac{1}{\cos \theta }$, we get

\[\begin{align}

  & LHS=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \dfrac{1}{\cos \theta }-\cos \theta \right) \\

 & \,\,\,\,\,\,\,\,\,\,=\left( \dfrac{1-\sin \theta \left( \sin \theta \right)}{\sin \theta } \right)\left( \dfrac{1-\cos \theta \left( \cos \theta \right)}{\cos \theta } \right) \\

 & \,\,\,\,\,\,\,\,\,\,=\left( \dfrac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \right) \\

\end{align}\]

As we know that $\begin{align}

  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\

 & \\

\end{align}$

Hence, $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $ and

              $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $

Using, $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $ and $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ in the above, expression, we get-

\[\] $\begin{align}

  & LHS=\left( \dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta } \right) \\

 & LHS=\left( \dfrac{{{\cos }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{{{\sin }^{2}}\theta }{\sin \theta } \right) \\

 & LHS=\left( \cos \theta \right)\left( \sin \theta \right).........................(i) \\

\end{align}$

Now, taking $RHS=\dfrac{1}{\tan \theta +\cot \theta }$

Using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }\,and\,\,\,\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ in the above expression, we get-$\begin{align}

  & RHS=\dfrac{1}{\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta }} \\

 & RHS=\dfrac{1}{\dfrac{\left( sin\theta \right)\left( \sin \theta \right)+\left( \cos \theta \right)\left( cos\theta \right)}{\left( \sin \theta \right)\left( \cos \theta \right)}} \\

 & RHS=\dfrac{1}{\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\left( \sin \theta \right)\left( \cos \theta \right)}} \\

 & RHS=\dfrac{1}{\dfrac{1}{\sin \theta \times \cos \theta }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \operatorname{Sin}ce,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right] \\

 & RHS=\sin \theta \times \cos \theta ...................(ii) \\

 & \\

\end{align}$

From equation (i) and (ii), we can conclude LHS=RHS

Hence, $\left( \sin \theta -\cos ec\theta \right)\left( \cos \theta -\sec \theta \right)=\dfrac{1}{\tan \theta +\cot \theta }(proved)$


Note: Students can also solve this question by this method, given as-
$\begin{align}

  & LHS=\left( \sin -\cos ec\theta \right)\left( \cos \theta -\sec \theta \right) \\

 & =\left( \sin \theta -\dfrac{1}{\sin \theta } \right)\left( \cos \theta -\dfrac{1}{\cos \theta } \right) \\

 & =\left( \dfrac{{{\sin }^{2}}\theta -1}{\sin \theta } \right)\left( \dfrac{{{\cos }^{2}}\theta -1}{\cos \theta } \right) \\

 & =\left( \dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta } \right) \\

 & LHS=\left( \dfrac{{{\cos }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{{{\sin }^{2}}\theta }{\sin \theta } \right) \\

 & LHS=\left( \cos \theta \right)\left( \sin \theta \right) \\

\end{align}$

Multiplying and divide by $\tan \theta +\cot \theta $ in the above expression, we get-
$\begin{align}

  & LHS=\left( \cos \theta \right)\left( \sin \theta \right)\times \dfrac{\left( tan\theta +\cot \theta \right)}{\left( \tan \theta +\cot \theta \right)} \\

 & LHS=\dfrac{\left( \sin \theta \right)\left( \cos \theta \right)}{\left( \tan \theta +\cot \theta \right)}\times \left( \dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta } \right) \\

 & LHS=\dfrac{\sin \theta \times \cos \theta }{\tan \theta +\cot \theta }\times \left( \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \times \cos \theta } \right) \\

 & LHS=\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\tan \theta +\cot \theta } \\

 & LHS=\dfrac{1}{\tan \theta +\cot \theta }\,\,\,\,\,\,\,\,\,\,\,\left( \because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right) \\

 & \\

\end{align}$

$\therefore $ LHS=RHS

Hence,

$\left( \sin \theta -\cos ec\theta \right)\left( \cos \theta -\sec \theta \right)=\dfrac{1}{\tan \theta +\cot \theta }(proved)$