Question

Prove that
$\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\sin A+\cos A$

Answer 1

- Hint:Convert the whole expression to $\sin \theta ,\cos \theta $ by replacing
 $\begin{align}
  & \tan \theta \Rightarrow \dfrac{\sin \theta }{\cos \theta }, \\
 & \cot \theta \Rightarrow \dfrac{\cos \theta }{\sin \theta } \\
\end{align}$
Hence, simplify the left hand side of the given equation and use the algebraic identity of ${{a}^{2}}-{{b}^{2}}$ which can be given as
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$

Complete step-by-step solution -

So, we have to prove
$\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\sin A+\cos A......................\left( i \right)$
As, we can prove the above relation by simplifying the left hand side of the equation and hence try to get the relation in the right hand side of the equation. So, we have LHS of the equation (i) as
$LHS=\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}........................\left( ii \right)$
Now, we can get the whole relation of LHS by replacing $\tan A,\cot A$ using the relation
$\tan \theta =\dfrac{\sin \theta }{\cos \theta },cot\theta =\dfrac{\cos \theta }{\sin \theta }.....................\left( iii \right)$
Now, we can get equation (i) as
$\begin{align}
  & LHS=\dfrac{\cos A}{1-\dfrac{\sin A}{\cos A}}+\dfrac{\sin A}{1-\dfrac{\cos A}{\sin A}} \\
 & LHS=\dfrac{\cos A}{\dfrac{\cos A-\sin A}{\cos A}}+\dfrac{\sin A}{\dfrac{\sin A-\cos A}{\operatorname{sinA}}} \\
 & LHS=\dfrac{\cos A\times \cos A}{\cos A-\cos A}+\dfrac{\sin A\times \sin A}{\sin A-\cos A} \\
\end{align}$


$LHS=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}+\dfrac{{{\sin }^{2}}A}{\sin A-\cos A}$
Now, taking ‘-‘ sign common from the second term of the above expression, we get
$\begin{align}
  & LHS=\dfrac{{{\cos }^{2}}A}{\cos A-\sin A}-\dfrac{{{\sin }^{2}}A}{\cos A-\sin A} \\
 & LHS=\dfrac{{{\cos }^{2}}A-{{\sin }^{2}}A}{\cos A-\sin A}..........................\left( iv \right) \\
\end{align}$
Now, we can use the algebraic identity of ${{a}^{2}}-{{b}^{2}}$ with the numerator of the equation (iv). So, identity is given as
${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)...................\left( v \right)$
Hence, we can re-write the numerator of the equation (iv) with the help of equation (v) as
$\begin{align}
  & LHS=\dfrac{{{\left( \cos A \right)}^{2}}-{{\left( \sin A \right)}^{2}}}{\cos A-\sin A} \\
 & LHS=\dfrac{\left( \cos A-\sin A \right)\left( \cos A+\sin A \right)}{\left( \cos A-\sin A \right)} \\
\end{align}$
Now, we can observe that the term $\cos A-\sin A$ is common to the numerator and denominator of the equation of LHS. So, we can cancel out the same term $\left( \cos A-\sin A \right)$ from the expression of LHS and hence, we get value of LHS as
$LHS=\cos A+\sin A,\sin A+\cos A....................\left( vi \right)$
Now, we can observe the simplified values of LHS of the equation (i) as $\cos A+\sin A$ which is the same as the value of RHS of the equation (i). Hence, we get
$LHS=RHS=\sin A+\cos A$
Hence, the given expression in the problem is proved.

Note: Converting the whole expression to $\sin \theta ,\cos \theta $ using relation
$\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
Is the key point of the problem. One may take LCM of $1-\tan \theta ,1-\cot \theta $ and solve them in terms of $\tan \theta ,\cot \theta $ and hence convert the result to $\sin \theta ,\cos \theta $ finally.
Don’t confuse the identities of $\tan \theta ,\cot \theta $ . As one may apply formulae of
$\tan \theta ,\cot \theta \Rightarrow \dfrac{\cos \theta }{\sin \theta },\dfrac{\sin \theta }{\cos \theta }$
Which are wrong and just opposite to each other. So, be clear with the relations of them. One may go wrong if he or she puts the root of the given problem as ‘-3’ by getting confused with the factor (x – 3). So, don’t confuse with the factor and root terminologies. Root of any factor can be calculated by equating the factor to 0.