" 15."cot A=(1)/(2)(cot(A)/(2)-tan(A)/(2))

If A+B+C=pi, show that : tan.(A)/(2)tan.(B)/(2)+tan.(B)/(2)tan.(C)/(2)+tan.(C)/(2)tan.(A)/(2)=1 Hence deduce that : cot.(A)/(2)+cot.(B)/(2)+.cot.(C)/(2)=cot.(A)/(2).cot.(B)/(2)tan.(C)/(2) .

In a Delta ABC, if cot((A)/(2)):cot((B)/(2)):cot((C)/(2))=1:4:15 then the greatest angle is

(cot A-1)/(2-sec^(2)A)=(cot A)/(1+tan A)

(cot A-1)/(2-sec^(2)A)=(cot A)/(1+tan A)

(cot A-1)/(2-sec^(2)A)=(cot A)/(1+tan A)

(cot A-1)/(2-sec^(2)A)=(cot A)/(1+tan A)

If A+B+C=pi , prove that (a) tan 3A+tan3B+tan3C=tan3A tan 3B tan 3C (b) cot""(A)/(2)+cot""(B)/(2)+cot""(C)/(2)=cot""(A)/(2)cot""(B)/(2)cot""(C)/(2)

The value of cot((15)/(2))^(@)+tan((135)/(2))^(@)-cot((135)/(2))^(@)-tan((15)/(2)) is

tan A-cot A=-2cot2A