If cosA=mcosB, then write the value of cotA+B2cotA−B2
We have.
cosA=mcosB,cosAcosB=m⇒cosAcosB+1=m+1⇒cosA+cosBcosB=m+1 …(i)
Again,
cosAcosB=m⇒cosAcosB−1=m−1⇒cosA−cosBcosB=m−1…(ii)
Dividing equation (i) by equation (ii), we get
cosA+cosBcosBcosA−cosBcosB=m−1m−1⇒cosA+cosBcsosA−cosB=fracm−1m−1⇒2cos(A+B2)cos(A−B2)2sin(A+B2)sin(A−B2)=m+1m−1⇒−cot(A+B2)cot(A−B2)=m+1m−1⇒cot(A+B2)cot(A−B2)=1+m1−m∴cot(A+B2)cot(A−B2)=1+m1−m