Question

show that cosA/ 1-tanA+sinA/1-cotA=cosA +sinA

Answer 1

Step-by-step explanation:

Hence proved

Yamansharma

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Answer 2

$\displaystyle \sf{ \frac{cos \:A }{1 - tan\:A} + \frac{sin\:A}{1 - cot\:A} = cos\:A + sin\:A} \: \: is \: \bf proved$

Given :

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

To find :

To prove the expression

Solution :

Step 1 of 2 :

Write down the given expression to prove

Here the given expression is

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

Step 2 of 2 :

Prove the expression

LHS

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

$\displaystyle \sf{ = \frac{cos \:A }{1 - \dfrac{sin\:A}{cos\:A} } + \frac{sin\:A}{1 - \dfrac{cos\:A}{sin\:A} } }$

$\displaystyle \sf{ = \frac{cos \:A }{ \dfrac{cos\:A - sin \:A}{cos\:A} } + \frac{sin\:A}{ \dfrac{sin\:A - cos\:A}{sin\:A} } }$

$\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{{sin}^{2} \:A}{sin\:A - cos\:A} }$

$\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} + \frac{{sin}^{2} \:A}{ - (cos\:A - sin \:A)} }$

$\displaystyle \sf{ = \frac{{cos}^{2} \:A }{ cos\:A - sin \:A} - \frac{{sin}^{2} \:A}{ cos\:A - sin \:A} }$

$\displaystyle \sf{ = \frac{{cos}^{2} \:A - {sin}^{2} \:A}{ cos\:A - sin \:A} }$

$\displaystyle \sf{ = \frac{(cos\:A + sin \:A)(cos\:A - sin \:A)}{ cos\:A - sin \:A} }$

$\displaystyle \sf{ = (cos\:A + sin \:A) }$

= RHS

Hence the proof follows

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