Question

Prove that :

$\frac{1 + { \tan}^{2} }{1 + \ \ { \cot }^{2} } = { \frac{1 - \tan \: }{1 - \cot } }^{2} = { \tan }^{2}$
​

Answer 1

CORECT QUESTION :-

$\\ \sf\: \dfrac{1 + {tan}^{2} \theta }{1 + {cot}^{2} \theta } = { \left( \dfrac{1 - tan \theta}{1 - cot \theta} \right)}^{2} = {tan}^{2} \theta$

SOLUTION :-

$\\ \sf \: (1) \: \dfrac{1 + {tan}^{2} \theta }{1 + {cot}^{2} \theta } \\ \\ \boxed{ \sf \: 1 + {tan}^{2} \theta = {sec}^{2} \theta } \: \: \: \: \: \: \: \: \: \boxed{ \sf \:1 + {cot}^{2} \theta = {sec}^{2} \theta }$

Putting values , we get...

$\\ \implies \sf \: \dfrac{ {sec}^{2} \theta}{ {cosec}^{2} \theta} \\ \\ \boxed{ \sf \: sec \theta = \dfrac{1}{cos \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \sf \: cosec \theta = \dfrac{1}{sin \theta} }$

Putting values , we get...

$\\ \sf \implies \: \dfrac{ \dfrac{1}{ {cos}^{2} \theta} }{ \dfrac{1}{ {sin}^{2} \theta } } \\ \\ \implies \sf \: \dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } \\ \\ \boxed{ \sf \: \frac{sin \theta}{cos \theta} = tan \theta }$

Putting values , we get ...

$\\ \implies \boxed{\boxed{ \sf \: {tan}^{2} \theta}} \: \: \: \: \: - - - (i)$

$\\ \sf \: (2) \: \: \: \: \: { \left( \dfrac{1 - {tan}^{} \theta }{1 - cot \theta} \right)}^{2} \\ \\ \boxed{ \sf \:tan \theta = \dfrac{sin \theta}{cos \theta} } \: \: \: \: \: \: \: \: \boxed{ \sf \: cot \theta= \dfrac{cos \theta}{sin \theta} }$

Putting values , we get...

$\\ \sf \implies { \left( \dfrac{1 - \dfrac{sin \theta }{cos \theta} }{1 - \dfrac{cos \theta}{sin \theta} } \right)}^{2} \\ \\ \\ \implies \sf { \left( \dfrac{ \dfrac{cos \theta - sin \theta}{cos \theta} }{ \dfrac{sin \theta- cos \theta}{sin \theta} } \right)}^{2}$

$\\ \implies \sf { \left( \dfrac{sin \theta(cos \theta - sin \theta)}{cos \theta(sin \theta - cos \theta)} \right)}^{2} \\ \\ \\ \implies \sf \dfrac{( {sin \theta)}^{2}( {cos \theta- sin \theta)}^{2} }{ {(cos \theta)}^{2} (sin \theta - {cos \theta)}^{2} } \\ \\ \\ \boxed{ \sf \: \dfrac{sin \theta}{cos \theta} = tan \theta }$

Putting values we get...

$\\ \implies \sf \: {tan}^{2} \theta \left( \dfrac{ {cos}^{2} \theta + {sin}^{2} \theta - 2sin \theta.cos \theta}{ {sin}^{2} \theta + {cos}^{2} \theta- 2sin \theta.cos \theta } \right) \\ \\ \\ \boxed{ \sf \: {sin}^{2} \theta + {cos}^{2} \theta = 1}$

Putting values we get...

$\\ \implies \sf \: {tan}^{2} \theta \cancel{\left( \dfrac{1 - 2sin \theta.cos \theta}{1 -2 sin \theta.cos \theta} \right)} \\ \\ \\ \implies \boxed{ \boxed{\sf \: {tan}^{2} \theta } }\: \: \: \: - - - - (ii)$

Hence , From equation (i) and (ii)..

$\\ \\ \underline{ \boxed{ \bf \: \dfrac{1 + {tan}^{2} \theta}{1 + {cot}^{2} \theta } = { \left( \dfrac{1 - tan \theta}{1 - cot \theta} \right)}^{2} = {tan}^{2} \theta }}$