Math, asked by sachin504091, 1 year ago

prove that (Sin5ACos2A-Sin6ACosA)/(SinASin2A-Cos2ACos3A)=tan A

Answers

Answered by kvnmurty

$\frac{Sin5ACos2A-Sin6ACosA}{SinASin2A-Cos2ACos3A}\\\\=\frac{[(Sin 7A+ Sin3A) - (Sin7A + SIn5A)]/2}{[(CosA - Cos3A)-(Cos5A+CosA)]/2}\\\\=-\frac{Sin3A-Sin5A}{Cos5A+COs3A}\\\\=\frac{Sin4A\ CosA+Cos4A\ SinA-Sin4A\ CosA+SinA Cos4A}{Cos4A\ CosA-SIn4A\ SinA+Cos4A\ CosA+Sin4A\ SinA}\\\\=tan A\\$

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