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Question: Prove that \[\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\dfrac{{1 - \tan ...

Prove that 1+tan⁡2θ1+cot⁡2θ=(1−tan⁡θ1−cot⁡θ)2=tan⁡2θ\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\dfrac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} = {\tan ^2}\theta .

Explanation

Solution

Here we are given to prove two different equalities altogether. We will take them one by one and then try to solve them. Using normal trigonometric formulas will help us to solve this problem like ,1+tan⁡2A=sec⁡2A1 + {\tan ^2}A = {\sec ^2}A, 1+cot2A=cosec2A1 + co{t^2}A = cose{c^2}A, and tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} .

Complete step-by-step answer:
Here we have,
1+tan⁡2θ1+cot⁡2θ=(1−tan⁡θ1−cot⁡θ)2=tan⁡2θ\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\dfrac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} = {\tan ^2}\theta
We start with,
(i) 1+tan⁡2θ1+cot⁡2θ=tan⁡2θ\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\tan ^2}\theta
L.H.S
1+tan⁡2θ1+cot⁡2θ\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }}
We now, have,
1+tan2A=sec2A1 + ta{n^2}A = se{c^2}A
1+cot2A=cosec2A1 + co{t^2}A = cose{c^2}A
On substituting the above values we get,
=sec2θcosec2θ= \dfrac{{se{c^2}\theta }}{{cose{c^2}\theta }}
Again,
sec2θ=1cos2θse{c^2}\theta = \dfrac{1}{{co{s^2}\theta }}
cosec2θ=1sin2θcose{c^2}\theta = \dfrac{1}{{si{n^2}\theta }}
So, again, if we continue,
=1cos2θ1sin2θ= \dfrac{{\dfrac{1}{{co{s^2}\theta }}}}{{\dfrac{1}{{si{n^2}\theta }}}}
=sin2θcos2θ= \dfrac{{si{n^2}\theta }}{{co{s^2}\theta }}
=tan2θ= ta{n^2}\theta
For the 2nd part,

(ii) we need to prove,
(1−tanθ1−cotθ)2=tan2θ{\left( {\dfrac{{1 - tan\theta }}{{1 - cot\theta }}} \right)^2} = ta{n^2}\theta
L.H.S
(1−tanθ1−cotθ)2{\left( {\dfrac{{1 - tan\theta }}{{1 - cot\theta }}} \right)^2}
As, tan⁡θ=sin⁡θcos⁡θ=1cot⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{1}{{\cot \theta }}
=(1−sin⁡θcos⁡θ1−cos⁡θsin⁡θ)2= {\left( {\dfrac{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}}} \right)^2}
On Simplifying, we get,
=(cosθ−sinθcos⁡θsinθ−cosθsin⁡θ)2= {\left( {\dfrac{{\dfrac{{cos\theta - sin\theta }}{{\cos \theta }}}}{{\dfrac{{sin\theta - cos\theta }}{{\sin \theta }}}}} \right)^2}
By cancelling out common terms we get,
=(1cos⁡θ−1sin⁡θ)2= {\left( {\dfrac{{\dfrac{1}{{\cos \theta }}}}{{\dfrac{{ - 1}}{{\sin \theta }}}}} \right)^2}
=(−sin⁡θcos⁡θ)2= {\left( {\dfrac{{ - \sin \theta }}{{\cos \theta }}} \right)^2}
As tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}, we get,
=(−tan⁡θ)2= {( - \tan \theta )^2}
On squaring we get,
=tan⁡2θ= {\tan ^2}\theta
=RHS
Hence, from (i) and (ii), we get 1+tan⁡2θ1+cot⁡2θ=(1−tan⁡θ1−cot⁡θ)2=tan⁡2θ\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\dfrac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} = {\tan ^2}\theta

Note: Some basic trigonometric functions had been used here in this problem. We had used 1+tan2A=sec2A1 + ta{n^2}A = se{c^2}Aand 1+cot2A=cosec2A1 + co{t^2}A = cose{c^2}A. Also we had deal with tan⁡θ=sin⁡θcos⁡θ=1cot⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{1}{{\cot \theta }}to simplify our problem and get our desired result.
One should remember all basic trigonometric properties and use them to simplify the problems.