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

Question

If tan A- tan B=x and cot B- cot A=y then cot (A-B) is (A) (1/x-y) (B) (1/x+y) (C) (1/x)+y (D) (1/x)+(1/y)

Tardigrade Admin · Accepted Answer

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/x) ∴ cot (A-B)=( cot A cot B+1/ cot B- cot A) =((y/x)+1/y)=(x+y/x y) =(1/x)+(1/y)

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

$\frac{1}{x - y}$

$\frac{1}{x + y}$

$\frac{1}{x} + y$

$\frac{1}{x} + \frac{1}{y}$

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

x−y1​

x+y1​

x1​+y

x1​+y1​

tanA−tanB=x ⇒cotA1​−cotB1​=x ⇒cotAcotBcotB−cotA​=x ⇒cotA⋅cotB=xy​ ∴cot(A−B)=cotB−cotAcotAcotB+1​ =yxy​+1​=xyx+y​ =x1​+y1​

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

$\frac{1}{x - y}$

$\frac{1}{x + y}$

$\frac{1}{x} + y$

$\frac{1}{x} + \frac{1}{y}$