If tan A- tan B=x and cot A- cot B=y then cot (A-B)=

Question

If tan A- tan B=x and cot A- cot B=y then cot (A-B)= (A) (x y/x+y) (B) (x y/x-y) (C) (x-y/x y) (D) (y-x/x y)

Tardigrade Admin · Accepted Answer

Since, tan A- tan B=x ⇒ (1/ cot A)-(1/ cot B)=x ⇒ ( cot B- cot A/ cot A cot B)=x ∵ cot A- cot B=y (given) So, cot A cot B=-(y/x) ∵ cot (A-B)=( cot A cot B+1/ cot B- cot A) =(-(y/x)+1/-y)=(y-x/x y)

Q. If $tan A - tan B = x$ and $cot A - cot B = y$ then $cot (A - B) =$

$\frac{x y}{x + y}$

$\frac{x y}{x - y}$

$\frac{x - y}{x y}$

$\frac{y - x}{x y}$

Q. If tanA−tanB=x and cotA−cotB=y then cot(A−B)=

x+yxy​

x−yxy​

xyx−y​

xyy−x​

Since, tanA−tanB=x ⇒cotA1​−cotB1​=x ⇒cotAcotBcotB−cotA​=x ∵cotA−cotB=y (given) So, cotAcotB=−xy​ ∵cot(A−B)=cotB−cotAcotAcotB+1​ =−y−xy​+1​=xyy−x​

Q. If $tan A - tan B = x$ and $cot A - cot B = y$ then $cot (A - B) =$

$\frac{x y}{x + y}$

$\frac{x y}{x - y}$

$\frac{x - y}{x y}$

$\frac{y - x}{x y}$