Question

prove that (cos A/1-tanA) - (sin²A/cosA-sinA) = cos A + sin A​

Answer 1

Step-by-step explanation:

$\sf\underline{To \: Prove:-}$

$\sf \dfrac{cosA}{1 - tanA} - \dfrac{ {sin}^{2}A }{cosA - sinA} = cosA + sinA$

$\sf\underline{Proof:-}$

$\sf LHS = \dfrac{cosA}{1 - tanA} - \dfrac{ {sin}^{2}A }{cosA - sinA}$

$\sf = \dfrac{cosA}{1 - \dfrac{sinA}{cosA} } - \dfrac{ {sin}^{2}A }{cosA - sinA}$

$\sf = \dfrac{cosA}{ \dfrac{</p><p>cosA - sinA }{cosA} } - \dfrac{ {sin}^{2}A }{cosA - sinA}$

$\sf = \dfrac{cosA \times cosA}{</p><p>cosA - sinA } - \dfrac{ {sin}^{2}A }{cosA - sinA}$

$\sf = \dfrac{cos ^{2} A}{</p><p>cos - sinA } - \dfrac{ {sin}^{2}A }{cosA - sinA}$

$\sf = \dfrac{cos ^{2} A}{</p><p>cosA - sinA } - \dfrac{ { sin}^{2}A }{ cosA - sinA}$

$\sf = \dfrac{cos ^{2} A - sin^{2}A }{</p><p>cosA -sinA }$

$\sf = \dfrac{(cos A - sinA) (cos A + sinA)}{</p><p>cosA -sinA }$

$\sf = cos A + sinA = RHS$

$\sf LHS = RHS$

$\sf \underline{Hence \: Proved}$