Question

Prove that:-
\[\dfrac{\cot A-\cos A}{\cot A+\cos A}=\dfrac{\operatorname{cosec}A-1}{\operatorname{cosec}A+1}\]

Answer 1

Hint: We proceed to prove the given equation starting from the expression at the left hand side . We convert the cotangent at the numerator and denominator of left hand side to sine and cosine using the identity $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ for a angle. We simplify until we get expression in sine angle in both numerator and denominator. We divide the numerator and denominator with $\sin A$ and use $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ to get the expression at the right hand side in $\operatorname{cosec}A$.\[\]

Complete step by step answer:
As mentioned in the question, we have to prove the given expression.
\[\dfrac{\cot A-\cos A}{\cot A+\cos A}=\dfrac{\operatorname{cosec}A-1}{\operatorname{cosec}A+1}\]
Now, we will start with the left hand side that is L.H.S. and try to make the necessary changes that are given in the hint, first, as follows
\[\begin{align}
  & =\dfrac{\cot A-\cos A}{\cot A+\cos A} \\
 & =\dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}+\cos A} \\
 & =\dfrac{\cos A-\sin A\cdot \cos A}{\cos A+\sin A\cdot \cos A} \\
\end{align}\]
Now, taking $\cos A$ common from both the numerator as well as the denominator, we can write as follows
\[\begin{align}
  & =\dfrac{\cos A-\sin A\cdot \cos A}{\cos A+\sin A\cdot \cos A} \\
 & =\dfrac{\cos A(1-\sin A)}{\cos A(1+\sin A)} \\
 & =\dfrac{(1-\sin A)}{(1+\sin A)} \\
\end{align}\]
Now, on dividing the numerator as well as the denominator with $\sin A$, we get the following result
\[=\dfrac{\left( \dfrac{1}{\sin A}-1 \right)}{\left( \dfrac{1}{\sin A}+1 \right)}\]
We know that $\operatorname{cosec} x = \dfrac{1}{{\sin x}}$ , hence, we can write as follows
\[=\dfrac{\left( \operatorname{cosec}A-1 \right)}{\left( \text{cosec}A+1 \right)}\]
Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved.\[\]

Note: We can alternatively solve starting from the right hand side of the statement equation and reach at the left hand side. We note that in the given statement of proof $\cot A$ and $\operatorname{cosec}A$ are well-defined which means $A\ne n\pi $ ($n$ as an integer) where the cotangent and cosec angles are not defined . That is why we could divide $\sin A$. We can similarly write the tangent and secant of an angle in terms of sine and cosine as $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\sec \theta =\dfrac{1}{\cos \theta }$ where $\theta \ne \dfrac{\left( 2n+1 \right)\pi }{2}$ for $n\in Z.$