Question

$\huge \tt Prove :$

$\sf \dfrac{ \cos A}{1- \tan A} + \dfrac{\sin A}{1-\cot A} = \cot A + \sin A$​

Answer 1

Answer:

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Answer 2

Correction in the question:

$\sf \dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA} = cosA + sinA$

LHS:

$\sf \Longrightarrow \dfrac{cosA}{1 - tanA} + \dfrac{sinA}{1 - cotA}$

We know that:

⇒ tanA = sinA/cosA

⇒ cotA = cosA/sinA

$\sf \Longrightarrow \dfrac{cosA}{1 - \Bigg\{\dfrac{sinA}{cosA}\Bigg\}} + \dfrac{sinA}{1 - \Bigg\{\dfrac{cosA}{sinA}\Bigg\}}$

$\sf \Longrightarrow \dfrac{cosA}{\Bigg\{\dfrac{cosA - sinA}{cosA}\Bigg\}} + \dfrac{sinA}{\Bigg\{\dfrac{sinA - cosA}{sinA}\Bigg\}}$

$\sf \Longrightarrow \Bigg\{cosA \times \dfrac{cosA}{cosA - sinA} \Bigg \} + \Bigg\{sinA \times \dfrac{sinA}{sinA - cosA} \Bigg \}$

$\sf \Longrightarrow\dfrac{cos^2A}{cosA - sinA} + \dfrac{sin^2A}{sinA - cosA}$

We know that:

⇒ (a - b) = -(b - a)

$\sf \Longrightarrow\dfrac{cos^2A}{cosA - sinA} - \dfrac{sin^2A}{cosA - sinA}$

$\sf \Longrightarrow\dfrac{cos^2A - sin^2A}{cosA - sinA}$

We know that:

⇒ a² - b² = (a - b)(a + b)

$\sf \Longrightarrow\dfrac{(cosA - sinA)(cosA + sinA)}{cosA - sinA}$

(cosA - sinA) & (cosA - sinA) in the numerator and denominator will get cancelled.

$\sf \Longrightarrow (cosA + sinA)$

Hence proved.