Question

1+ cos A / sin A - sin A / 1 + cos A = cot A​

Answer 1

Step-by-step explanation:

your question is wrong. in R.H.S it should be 2cot A. but still I have corrected the question and solved which is in the attachment paper.

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Answer 2

To Prove :

$\\\implies \sf \dfrac{1 + \cos A}{ \sin A} + \dfrac{ \sin A}{1 + \cos A} = 2\cot A$

Proof :

On Solving LHS

$\implies \sf \dfrac{1 + \cos A}{ \sin A} - \dfrac{ \sin A}{1 + \cos A} \\ \\ \implies \sf \dfrac{ {(1 + \cos A)}^{2} - {( \sin)}^{2} A}{( \sin A)(1 + \cos A)} \\ \\ \implies \sf \dfrac{ {1 +2 \cos A + \cos}^{2}A - {\sin}^{2}A }{( \sin A)(1 + \cos A)} \\ \\ \implies \sf \dfrac{ {2 \cos A + \cos}^{2}A + { \cos}^{2} A}{( \sin A)(1 + \cos A)} \\ \\ \implies \sf \dfrac{ {2 \cos A + 2\cos}^{2}A}{( \sin A)(1 + \cos A)} \\ \\ \implies \sf \dfrac{2 \cos A(1 + \cos A)}{( \sin A)(1 + \cos A)} \\ \\ \implies \sf \dfrac{2 \cos A}{ \sin A} \\ \\\large\implies \boxed{\boxed{\sf 2 \cot A}}$

LHS = RHS

Hence proved

Identity used :

$\\\large\sf \implies 1 - \sin^2 A = \cos^2 A \\ \\\large\sf \implies \dfrac{\cos A}{\sin A}= \cot A$