QUESTION

# Prove that, $\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}=2+\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}$.

Hint: Shift both the terms that contains trigonometric function on one side and then prove: $\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}-\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}=2$. First rationalize the denominator of both the terms and then use trigonometric identities to simplify the expression.

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 to prove: $\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}=2+\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}$ or we can also prove that,

$\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}-\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}=2$.

Therefore,

$L.H.S=$ $\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}-\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}$

Rationalizing the denominator we get,

\begin{align} & L.H.S=\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}\times \dfrac{\left( \cot A-\cos ecA \right)}{\left( \cot A-\cos ecA \right)}-\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}\times \dfrac{\left( \cot A+\cos ecA \right)}{\left( \cot A+\cos ecA \right)} \\ & =\dfrac{\sin A\cot A-\sin A\cos ecA}{\left( \cot A+\cos ecA \right)\left( \cot A-\cos ecA \right)}-\dfrac{\sin A\cot A+\sin A\cos ecA}{\left( \cot A+\cos ecA \right)\left( \cot A-\cos ecA \right)} \\ \end{align}

Using the identity: $\sin A\cos ecA=1\text{ and }(a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ we get,
$L.H.S=\dfrac{\sin A\cot A-1}{\left( {{\cot }^{2}}A-\cos e{{c}^{2}}A \right)}-\dfrac{\sin A\cot A+1}{\left( {{\cot }^{2}}A-\cos e{{c}^{2}}A \right)}$

Applying the identity given by, $\cos e{{c}^{2}}A-{{\cot }^{2}}A=1$, we get,

\begin{align} & L.H.S=\left( \dfrac{\sin A\cot A-1}{-1} \right)-\left( \dfrac{\sin A\cot A+1}{-1} \right) \\ & =(1-\sin A\cot A)+(\sin A\cot A+1) \\ & =2 \\ & =R.H.S \\ \end{align}

Therefore, we have proved that, $\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}-\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}=2$.

Hence, again shifting the trigonometric term to its original position we get, $\dfrac{\sin A}{\left( \cot A+\cos ecA \right)}=2+\dfrac{\sin A}{\left( \cot A-\cos ecA \right)}$.

Note: It is important to note that this type of approach can be very useful in proving such types of questions which are difficult to prove from one side. An alternate method can be applied. We can start simplifying the left hand side term and we will stop when it cannot be further simplified. Now, we will start simplifying the right hand side term and simplify it till we get it equal to the simplified left hand side term.