Math, asked by shubhanshu85, 9 months ago

sin (A - B)
cos A cos B
sin (B-C)
cos B cos C
sin (C - A)
cos C cos A
= 0​

Answers

Answered by arkaroy45
1

Step-by-step explanation:

Given, sin(a-b)/cos a cos b + sin(b-c)/cos b cos c + sin(c-a)/cos c cos a

Given, sin(a-b)/cos a cos b + sin(b-c)/cos b cos c + sin(c-a)/cos c cos a

= sin a cos b - cos a sin b/cos a cos b + sin b cos c - cos b sin c/cos b cos c + sin c cos a - cos c sin a/cos c cos a

= sin a cos b/cos a cos b - cos a sin b/cos a cos b + sin b cos c/cos b cos c - cos b sinc/cos b cos c + sin c cos a/cos c cos a - cos c sin a/cos c cos a

= tan a - tan b + tan b - tan c + tan c - tan a= 0.


shubhanshu85: are bhai full solution karke batao
arkaroy45: okk
arkaroy45: can you recheck the math
arkaroy45: i think the math should be like this.... Sin(A-B)/cosA.cosB + sin (B-C)/cosB.cosC +sin (C-A)/cosC.cosA =0
shubhanshu85: ha यही h bhai
shubhanshu85: plzz solve this
arkaroy45: okkk
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