Question

prove thatcot A tan(180+A)+tan(90+A)+tan(360-A)=0

Answer 1

L.H.S

CotA + tan(180+A) + tan(90+A)+tan(360-A)

= CotA + tan(2*90+A) + tan(90+A) + tan(4*90-A)

= CotA + tanA - CotA - tanA

= 0 = R.H.S proved

Answer 2

Answer:

$cot A tan(180+A)+tan(90+A)+tan(360-A)=0\ It\ is\ proved$

Step-by-step explanation:

Given: Given expression

To find: cot A tan(180+A)+tan(90+A)+tan(360-A)=0

Solution:

tan (90°+a) = - cot a,

sin (180°-a) = sin a,

cos (90°-a) = sin a,

cos (360°-a) = cos a

$cot A\ tan(180+A)+tan(90+A)+tan(360-A)=0$

$cot A\ tan A-cot A +tan(360-A)=0 \Rightarrow tan (90°+a) = - cot a, tan (180°+a) = tan a,\\cot A\ tan A-cot A +tan A =0 \Rightarrow tan (90°+a) = - cot a, tan (180°+a) = tan A,\\L.H.S=R.H.S$