Question

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

Answer 1

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-$\Rightarrow \dfrac{{\dfrac{{\sin A}}{{\cos A}}}}{{1 + \dfrac{{\cos A}}{{\sin A}}}} + \dfrac{{\dfrac{{\cos A}}{{\sin A}}}}{{1 + \dfrac{{sinA}}{{co\operatorname{s} A}}}}$

$$\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 –

$$\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