Question

cosA/1-tanA+sinA/1-cotA=sinA+CosA​

Answer 1

Answer:

You have not given the exact question.....

Of we have to prove this then your required answer is in the image attached......

Step-by-step explanation:

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And one thing I've ommitted some steps...

I hope you'll understand it easily.....ʘ‿ʘ

Answer 2

Solution :

$\implies \sf\frac{ \cos A}{1 - \tan A} + \frac{ \sin A}{1 - \cot A} \\ \\\implies \sf \frac{ \cos A}{1 - \frac{ \sin A}{ \cos A} } + \frac{ \sin A}{1 - \frac{ \cos A}{ \sin A} } \\ \\ \implies \sf \frac{ \cos A}{ \frac{ \cos A- \sin A}{ \cos A} } + \frac{ \sin A}{ \frac{ \sin A- \cos A}{ \sin A} } \\ \\\implies \sf \frac{ { \cos}^{2} A}{ \cos A- \sin A} + \frac{ { \sin}^{2} A}{ \sin A- \cos A} \\ \\\implies \sf \frac{ { \cos}^{2} A}{ \cos - \sin A} - \frac{ { \sin}^{2}A }{ \cos A- \sin A} \\ \\\implies \sf \frac{ { \cos}^{2}A - { \sin}^{2}A }{ \cos A - \sin A} \\ \\ \implies \sf\frac{( \cos A+ \sin A)( \cos A- \sin A)}{ \cos A- \sin A} \\ \\ \implies \sf\sin A + \cos A$

LHS = RHS

Hence Proved

Identity used :

$\implies\sf \tan A = \dfrac{\sin A}{\cos A} \\ \\ \sf \implies a^2 - b^2 = (a+b)(a-b)$