Question

Q. Show that .

[ cos^2a / ( 1 - tan a )] + [ sin^3a / ( sin a - cos a ) ] = 1 + sin a × cos a.



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Answer 1

$Given : \frac{cos^2A}{1 - tanA} + \frac{sin^3A}{sinA - cosA}$

$= \ \textgreater \ \frac{cos^2A}{1 - \frac{sinA}{cosA} } + \frac{sin^3A}{sinA - cosA}$

$= \ \textgreater \ \frac{cos^2A}{ \frac{cosA - sinA}{cosA} } + \frac{sin^3A}{sinA-cosA}$

$= \ \textgreater \ \frac{cos^2A * cosA}{cosA - sinA} + \frac{sin^3A}{sinA - cosA}$

$= \ \textgreater \ \frac{cos^3A}{cosA - sinA} + \frac{sin^3A}{sinA - cosA}$

$= \ \textgreater \ \frac{sin^3A - cos^3A}{sinA - cosA}$

$= \ \textgreater \ \frac{(sinA - cosA)(sin^2A + cos^2A + sinAcosA)}{sinA - cosA}$

$= \ \textgreater \ sin^2A + cos^2A + sinAcosA$

= > 1 + sinAcosA


Hope this helps!

Answer 2

Hi,

Please see the attached file!


Thanks