Question

1 + tan^2A/ 1 + cot^2A = (1 - tan^2/1 - cot^2A)^2 = tan^2 A .Prove this

Answer 1

Question : -

Prove that : -

1 + tan² A/1 + cot² A = ( 1 - tan A/1 - cot A )² = tan² A .

ANSWER

Given : -

1 + tan² A/1 + cot² A = ( 1 - tan A/1 - cot A )² = tan² A .

Required to prove : -

• 1 + tan² A/1 + cot² A = tan² A

• ( 1 - tan A/1 - cot A )² = tan² A

Identities used : -

sin² A + cos² A = 1

Proof : -

1 + tan² A/1 + cot² A = ( 1 - tan A/1 - cot A )² = tan² A .

We need to prove that ;

1 + tan² A/1 + cot² A = tan² A

( 1 - tan A/1 - cot A )² = tan² A

So,

Let's consider the 1st part : -

$\sf \dfrac{1 + { \tan}^{2} A }{1 + { \cot}^{2}A } = { \tan}^{2} A \\ \\ \sf Consider \: the \: LHS \: part \: \\ \\ \sf \dfrac{1 + { \tan }^{2} A}{ 1 + {\cot}^{2} A } \\ \\ \mathsf{ Since, \tan \theta = \dfrac{ \sin \theta }{ \cos \theta } \quad ( and ) \quad \cot \theta = \dfrac{ \cos \theta }{ \sin \theta}} \\ \\ \dfrac{ 1 + \dfrac{ { \sin}^{2} A}{ {\cos}^{2} A }}{1 + \dfrac{ {\cos }^{2} A }{ {\sin}^{2} A }}$

$\sf\dfrac{ \dfrac{ {\cos}^{2} A + {\sin}^{2} A }{ {\cos}^{2} A } }{ \dfrac{ {\sin}^{2} A+ {\cos}^{2}A }{ {\sin}^{2} A} } \\ \\ \\ \sf Since , We \: know \: that \\ {\sin}^{2} + {\cos}^{2} = 1 \\ \\ \\ \tt \dfrac{ \dfrac{1}{ {\cos}^{2}A } }{ \dfrac{1}{ {\sin}^{2} A } } \\ \\ \sf \dfrac{ 1 }{ \cos^2 A } \times \dfrac{ \sin^2 A }{ 1 } \\ \\ \sf \dfrac{ \sin^2 A }{ \cos^2 A} \\ \\ \sf \implies {\tan}^{2} A$

Similarly,

Consider the 2nd part : -

$\displaystyle \sf \left( \dfrac{1 - \tan A }{ 1 - \cot A }\right)^2 = \tan^2 A\\ \\ \sf Consider \ the \ LHS \ part ; \\ \\ \sf \left( \dfrac{ 1 - \tan A }{ 1 - \cot A }\right)^2 \\ \\ \\ \sf We \ know \ that, \\ \tt \tan A = \dfrac{ \sin A}{ \cos A} \\ \\ \\ \sf \left( \dfrac{ 1 - \dfrac{ \sin A }{ \cos A }}{ 1 - \dfrac{ \cos A }{ \sin A }} \right )^2 \\ \\ \\ \sf \left( \dfrac{ \left[ 1 - \dfrac{ \sin A}{ \cos A} \right]^2}{\left[ 1 - \dfrac{ \cos A}{\sin A} \right]^2} \right) \\ \\ \sf Using \ the \ identities \\ \rm \bullet ( x - y )^2 = x^2 + y^2 - 2xy \\ \\ \\ \left( \dfrac{ \left[ ( 1 )^2 + \bigg(\dfrac{ \sin A }{ \cos A }\bigg)^2 - 2 ( 1 )\bigg( \dfrac{ \sin A }{ \cos A } \bigg)\right]}{ \left[ ( 1 )^2 + \bigg( \dfrac{ \cos A }{ \sin A } \bigg)^2 - 2 ( 1 ) \bigg( \dfrac{ \cos A }{ \sin A }\bigg) \right] }\right)$

$\displaystyle \sf \left( \dfrac{ 1 +\dfrac{ \sin^2 A}{ \cos^2 A} - \dfrac{ 2 \sin A}{ \cos A}}{1 + \dfrac{ \cos^2 A}{ \sin^2 A} - \dfrac{2 \cos A}{ \sin A}} \right) \\ \\ \sf By \; taking \; the \; respective \; LCM's \\ \\ \\ \sf \left( \dfrac{ \dfrac{ \cos^2 A + \sin^2A - 2 \sin A \cos A}{ \cos^2 A}}{ \dfrac{ \sin^2 A + \cos^2 A - 2 \cos A \sin A }{ \sin^2 A}}\right) \\ \\ \textsf{The numerator of each fraction get's cancelled because they are alike } \\ \\ \\ \sf \left( \dfrac{ \dfrac{ 1}{ \cos^2 A}}{ \dfrac{1}{ \sin^2 A}} \right) \\ \\ \sf \dfrac{1}{ \cos^2 A} \times \dfrac{\sin^2 A}{1} \\ \\ \sf \dfrac{ \sin^2 A}{ \cos^2 A } \\ \\ \tt \implies \tan^2 A$

LHS = RHS

Hence Proved !