Question

1+Sin (A-B)/cosAcosB+sin(B-C)/cosBcosC+sin(C-A)/cosCcosA

Answer 1

Answer:

That line becomes [[sinBcosC - cosBsinC)/cosBcosC] + [(sinCcosA = cosCsinA)/cosAcosC] + [(sinAcosB - cosAsinB)/cosAcosB] and if tyou look closely you will see that in each bracket the denominator cancels with the numerator, and we are left with sinB/cosB - sinC/cosc + sinC/cosC - sinA/cosA + sinA/cosA - sinB/cosB and if you cannot see that is 0, continue with the fact that sinC/cosC = tan C etc so we get tanB - tanC + tanC - tanA + tanA - tanB and as you can see that is 0

Step-by-step explanation:

Sir your question is different from this but method is same

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