Question

Prove that, $\dfrac{{\cos A.\cot A}}{{1 - \sin A}} = 1 + \cos ecA$ .

Answer 1

In this question, there is a trigonometric equation. But the task is not to find the value of the concerned angle $A$, but to prove both sides of the above equation to be equal. Again, a trigonometric function is a real valued function which has an angle of a right-angled triangle when the two side lengths of the triangle are divided. Here, $\cos ,$ $\cot$, $\sin$ and $\cos ec$ are all trigonometric functions.

Complete step-by-step solution:
Here is to prove that $\dfrac{{\cos A.\cot A}}{{1 - \sin A}} = 1 + \cos ecA$.
Now,
Left Hand Side, L.H.S.,
$\dfrac{{\cos A.\cot A}}{{1 - \sin A}} = \dfrac{{\cos A.\left( {\dfrac{{\cos A}}{{\sin A}}} \right)}}{{1 - \sin A}}$ [ as $\cot A = \dfrac{{\cos A}}{{\sin A}}$]
$= \dfrac{{{{\cos }^2}A}}{{\sin A\left( {1 - \sin A} \right)}}$ … … …(i)
But we know, ${\sin ^2}A + {\cos ^2}A = 1$ i.e., ${\cos ^2}A = 1 - {\sin ^2}A$.
Then putting the value of ${\cos ^2}A = 1 - {\sin ^2}A$ in (i), we obtain $= \dfrac{{1 - {{\sin }^2}A}}{{\sin A\left( {1 - \sin A} \right)}} = \dfrac{{\left( {1 + \sin A} \right)\left( {1 - \sin A} \right)}}{{\sin A\left( {1 - \sin A} \right)}}$.
As because, ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$, then $1 - {\sin ^2}A = \left( {1 + \sin A} \right)\left( {1 - \sin A} \right)$ .
Therefore, $= \dfrac{{1 + \sin A}}{{\sin A}} = \dfrac{1}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} = \cos ecA + 1$ =Right Hand Side[R.H.S.]
Therefore, $\dfrac{{\cos A.\cot A}}{{1 - \sin A}} = 1 + \cos ecA$. i.e., L.H.S.=R.H.S.

Note: Students should note that the easiest way to solve these types of problems is to use the normal trigonometric formulae. In this particular problem, we have used the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$. There are few more formulae for the relations in trigonometric functions. Any kind of trigonometric problem can be evaluated or proved in this method.