Question

Prove that \[\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A\cos ecA\]

Answer 1

Hint: In left hand side convert all \[\tan A\] and \[\cot A\] in the form of \[\sin A\] & \[ \cos A\] then try to take the lcm of the terms left after simplifying it, you will observe its all coming down in the form of a formula after putting it you can easily get the answer.

Complete step-by-step answer:
We know that \[\tan A = \dfrac{{\sin A}}{{\cos A}}\] & \[ \cot A = \dfrac{{\cos A}}{{\sin A}}\]
By putting this in the Left hand side we will try to find the right hand side on putting the above mention values of \[\tan A\] & \[ \cot A\] in the left hand side we will get it as
\[\begin{array}{l}
 \Rightarrow \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{\dfrac{{\cos A - \sin A}}{{\cos A}}}}\\
 \Rightarrow \dfrac{{{{\sin }^2}A}}{{\cos A(\sin A - \cos A)}} + \dfrac{{{{\cos }^2}A}}{{\sin A(\cos A - \sin A)}}
\end{array}\]

Now take a negative common from the denominator of the second term i.e.,
 \[(\cos A - \sin A)\] which will make it look like \[ - (\sin A - \cos A)\]
Now if we multiply the + in front of the term with the - in the denominator it will look like
\[\begin{array}{*{20}{l}}
{}\\
{ \Rightarrow \dfrac{{{{\sin }^2}A}}{{\cos A(\sin A - \cos A)}} - \dfrac{{{{\cos }^2}A}}{{\sin A(\sin A - \cos A)}}}
\end{array}\]

Now simplifying it further by taking the lcm we will get it as
\[\begin{array}{*{20}{l}}
{}\\
{ \Rightarrow \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{\cos A\sin A(\sin A - \cos A)}}}
\end{array}\]

Now apply the formula \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\]
Now put \[a = \sin A\] & \[ b = \cos A\] then we can rewrite
\[ \Rightarrow {\sin ^3}A - {\cos ^3}A = (\sin A - \cos A)({\sin ^2}A + \sin A\cos A + {\cos ^2}A)\]
Putting this in the last step
\[\begin{array}{l}
 \Rightarrow \dfrac{{(\sin A - \cos A)({{\sin }^2}A + \sin A\cos A + {{\cos }^2}A)}}{{\sin A\cos A(\sin A - \cos A)}}\\
 \Rightarrow \dfrac{{({{\sin }^2}A + \sin A\cos A + {{\cos }^2}A)}}{{\sin A\cos A}}\\

Now,\\
{\sin ^2}A + {\cos ^2}A = 1
\end{array}\]
So we are left with
\[\begin{array}{l}
 \Rightarrow \dfrac{{(\sin A - \cos A)({{\sin }^2}A + \sin A\cos A + {{\cos }^2}A)}}{{\sin A\cos A(\sin A - \cos A)}}\\
 \Rightarrow \dfrac{{1 + \sin A\cos A}}{{\sin A\cos A}}\\
 \Rightarrow \dfrac{{\sin A\cos A}}{{\sin A\cos A}} + \dfrac{1}{{\sin A\cos A}}\\
 \Rightarrow 1 + \dfrac{1}{{\sin A\cos A}}
\end{array}\]

Now we know that \[\dfrac{1}{{\sin A}} = \cos ecA\] & \[ \dfrac{1}{{\cos A}} = \sec A\]
Putting this we get the final answer as
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A\cos ecA\]
Hence Proved.


Note: Taking a negative sign out of the bracket from the denominator and multiplying with the positive in front of it before doing the lcm was the key step, if we don't do that the calculation will itself become too long and the possibilities of making a mistake in a long solution is very much likely.