Question

prove the following ​

Answer 1

Given : cotA/(1 - tanA)  +  tanA/(1 - cotA)  = 1 + tanA + cotA

To Find : Prove

Solution:

LHS =  cotA/(1 - tanA)  +  tanA/(1 - cotA)

use cotA = cosA/sinA   and tanA  = sinA/CosA

= (cosA/sinA)/(1 - sinA/cosA)   +  (sinA/CosA)/(1  - cosA/sinA)

=  (cosA/sinA)/((cosA - sinA)/cosA)   +  (sinA/CosA)/()sinA - cosA)/sinA)

= cos²A/sinA(cosA - sinA)   + sin²A/cosA(sinA - cosA)

= cos²A/sinA(cosA - sinA)   - sin²A/cosA(cosA - sinA)

=  (cos³A - sin³A)/sinAcosA(cosA - sinA)

using identity a³ - b³  = (a - b)(a²  + b²  + ab)

= (cosA - sinA)( cos²A + sin²A + cosAsinA)/ sinAcosA(cosA - sinA)

= ( cos²A + sin²A + cosAsinA) /  sinAcosA

= cos²A / sinAcosA +  sin²A / sinAcosA +  cosAsinA/sinAcosA

= cos A / sinA  +  sin A / cosA +  1

= cotA  + tanA + 1

=  1 + tanA + cotA

= RHS

QED

Hence proved

cotA/(1 - tanA)  +  tanA/(1 - cotA)  = 1 + tanA + cotA

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Answer 2

SOLUTION

TO PROVE

$\displaystyle \sf{ \frac{ \cot A}{1 - \tan A} + \frac{ \tan A}{1 - \cot A} =1 + \tan A + \cot A }$

PROOF

LHS

$\displaystyle \sf{ = \frac{ \cot A}{1 - \tan A} + \frac{ \tan A}{1 - \cot A} }$

$\displaystyle \sf{ = \frac{ \cot A}{1 - \tan A} + \frac{ \tan A}{1 - \displaystyle \sf{\frac{1}{ \tan A} } }}$

$\displaystyle \sf{ = \frac{ \cot A}{1 - \tan A} + \frac{ \tan A}{ \displaystyle \sf{\frac{\tan A - 1}{ \tan A} } }}$

$\displaystyle \sf{ = \frac{ \cot A}{1 - \tan A} + \frac{ {\tan}^{2} A}{ \tan A - 1}}$

$\displaystyle \sf{ = \frac{ \cot A}{1 - \tan A} - \frac{ {\tan}^{2} A}{ 1- \tan A}}$

$\displaystyle \sf{ = \frac{ \cot A - {\tan}^{2} A}{1 - \tan A} }$

$\displaystyle \sf{ = \frac{ \cot A -1 + 1 - {\tan}^{2} A}{1 - \tan A} }$

$\displaystyle \sf{ = \frac{ \cot A - \cot A. \tan A+ 1 - {\tan}^{2} A}{1 - \tan A} }$

$\displaystyle \sf{ = \frac{ \cot A (1- \tan A)+ (1 - {\tan}^{2} A)}{1 - \tan A} }$

$\displaystyle \sf{ = \frac{ \cot A (1- \tan A)}{1 - \tan A} + \frac{ (1 - {\tan}^{2} A)}{1 - \tan A}}$

$\displaystyle \sf{ = \cot A + \frac{ (1 + \tan A)(1 - \tan A)}{1 - \tan A}}$

$\displaystyle \sf{ = \cot A + 1 + \tan A}$

$\displaystyle \sf{ = 1 + \tan A + \cot A}$

= RHS

Hence proved

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