Question

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

Answer 1

Hint: Here we are given to prove a trigonometric equation that is to prove \[\dfrac{{\tan A}}{{1 + \cot A}} + \dfrac{{\cot A}}{{1 + \tan A}}\] is equal to the \[1 + \sec A\cos ecA\] so while solving such kind of question we would first break the \[\tan A\] into its components of \[\sin A\] and \[co\operatorname{s} A\] also same thing is done with the \[\cot A\] is to break it into its components of \[\sin A\] and \[co\operatorname{s} A\] later on solving the resultant equation to get the desired answer let us see the implementation of this concept and the step by step solution

Complete step-by-step answer:
Here are given with the things in the question that are to prove –
\[\dfrac{{\tan A}}{{1 + \cot A}} + \dfrac{{\cot A}}{{1 + \tan A}} = 1 + \sec A\cos ecA\]
Taking left hand side that is \[\dfrac{{\tan A}}{{1 + \cot A}} + \dfrac{{\cot A}}{{1 + \tan A}}\]
So first of all we need to simplify the above trigonometric equation to begin with and to get the desired result
For simplification we need to break down \[\tan A\] into its components of \[\sin A\] and \[co\operatorname{s} A\]that is
\[\tan A = \dfrac{{\sin A}}{{\cos A}}\] as \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] in a triangle as it is a basic identity of trigonometry
Also repeating the same stuff with the \[\cot A\] that is to break it into its components of \[\sin A\]and \[co\operatorname{s} A\] that becomes \[\cot A = \dfrac{{co\operatorname{s} A}}{{\sin A}}\] as \[co\operatorname{t} \theta = \dfrac{{co\operatorname{s} \theta }}{{\sin \theta }}\] in a triangle as it is a basic identity of trigonometry
Now putting the simplified values of \[\tan A\]and \[\cot A\] in the equation \[\dfrac{{\tan A}}{{1 + \cot A}} + \dfrac{{\cot A}}{{1 + \tan A}}\]
It becomes-\[\Rightarrow \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{1 + \dfrac{{\cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{1 + \dfrac{{sinA}}{{co\operatorname{s} A}}}}\]
\[ \Rightarrow
  \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{\dfrac{{\sin A - \cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{\dfrac{{\cos A - \sin A}}{{\cos A}}}} \\
  \dfrac{{{{\sin }^2}A}}{{\cos A\left( {\sin A - \cos A} \right)}} - \dfrac{{{{\cos }^2}A}}{{\sin A(\sin A - \cos A)}} \\
    \\
 \]
Now further solving the equation we get –
\[ \Rightarrow
  \dfrac{{{{\sin }^3}A}}{{\cos A\sin A\left( {\sin A - \cos A} \right)}} - \dfrac{{{{\cos }^3}A}}{{\cos A\sin A(\sin A - \cos A)}} \\
    \\
 \]
\[ \Rightarrow \dfrac{{{{\sin }^3}A - {{\cos }^3}A}}{{\cos A\sin A\left( {\sin A - \cos A} \right)}}\]
Now using the identity of the \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\] on the numerator of the above equation that is \[{\sin ^3}A - {\cos ^3}A\] the above equation becomes -
\[ \Rightarrow \dfrac{{(\sin A - \cos A)({{\sin }^2}A + \sin AcosA + {{\cos }^2}A)}}{{\cos A\sin A\left( {\sin A - \cos A} \right)}}\]
Cancelling out the like terms from the numerator and denominator we get -
\[ \Rightarrow \dfrac{{({{\sin }^2}A + \sin AcosA + {{\cos }^2}A)}}{{\cos A\sin A}}\]
Now using a basic trigonometric identity that is \[{\sin ^2} + {\cos ^2}A = 1\] in above equation we get –
\[\Rightarrow \dfrac{{1 + \sin AcosA}}{{\cos A\sin A}}\]
On further simplifying the above equation we get -
\[\Rightarrow \sec A\cos ecA + 1\](as \[\dfrac{1}{{\cos A}} = \sec A\] and \[\dfrac{1}{{sinA}} = \cos ecA\] is a basic trigonometric relation)
Hence the left hand side is equal to \[\sec A\cos ecA + 1\] which is also equal to the right hand as given in the question
Hence proved


Note: While solving such kind of questions care should be taken during solving the numerator and denominator as a mistake their can lead our whole question being wrong also one should be having the knowledge of the identities as they are crucial in solving such kind of questions