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# Prove that $\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\dfrac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} = {\tan ^2}\theta$.

Last updated date: 04th Aug 2024
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Hint: 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 ,$1 + {\tan ^2}A = {\sec ^2}A$, $1 + co{t^2}A = cose{c^2}A$, and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ .

Here we have,
$\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\dfrac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} = {\tan ^2}\theta$
(i) $\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\tan ^2}\theta$
L.H.S
$\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }}$
We now, have,
$1 + ta{n^2}A = se{c^2}A$
$1 + co{t^2}A = cose{c^2}A$
On substituting the above values we get,
$= \dfrac{{se{c^2}\theta }}{{cose{c^2}\theta }}$
Again,
$se{c^2}\theta = \dfrac{1}{{co{s^2}\theta }}$
$cose{c^2}\theta = \dfrac{1}{{si{n^2}\theta }}$
So, again, if we continue,
$= \dfrac{{\dfrac{1}{{co{s^2}\theta }}}}{{\dfrac{1}{{si{n^2}\theta }}}}$
$= \dfrac{{si{n^2}\theta }}{{co{s^2}\theta }}$
$= ta{n^2}\theta$
For the 2nd part,

(ii) we need to prove,
${\left( {\dfrac{{1 - tan\theta }}{{1 - cot\theta }}} \right)^2} = ta{n^2}\theta$
L.H.S
${\left( {\dfrac{{1 - tan\theta }}{{1 - cot\theta }}} \right)^2}$
As, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{1}{{\cot \theta }}$
$= {\left( {\dfrac{{1 - \dfrac{{\sin \theta }}{{\cos \theta }}}}{{1 - \dfrac{{\cos \theta }}{{\sin \theta }}}}} \right)^2}$
On Simplifying, we get,
$= {\left( {\dfrac{{\dfrac{{cos\theta - sin\theta }}{{\cos \theta }}}}{{\dfrac{{sin\theta - cos\theta }}{{\sin \theta }}}}} \right)^2}$
By cancelling out common terms we get,
$= {\left( {\dfrac{{\dfrac{1}{{\cos \theta }}}}{{\dfrac{{ - 1}}{{\sin \theta }}}}} \right)^2}$
$= {\left( {\dfrac{{ - \sin \theta }}{{\cos \theta }}} \right)^2}$
As $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$, we get,
$= {( - \tan \theta )^2}$
On squaring we get,
$= {\tan ^2}\theta$
=RHS
Hence, from (i) and (ii), we get $\dfrac{{1 + {{\tan }^2}\theta }}{{1 + {{\cot }^2}\theta }} = {\left( {\dfrac{{1 - \tan \theta }}{{1 - \cot \theta }}} \right)^2} = {\tan ^2}\theta$

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