Question

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

Answer 1

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

Explanation:

Given trigonometric equations:

`\begin{cases}\tan A - \tan B = x\\\cot B - \cot A = y\end{cases}`

`\textsf{Using the identity\;\;$\cot \theta=(1)/(\tan \theta),$\;\;rewrite the second equation in terms of $\tan$:}`

`\begin{aligned}y&=\cot B - \cot A\\\\y&=(1)/(\tan B) - (1)/(\tan A)\\\\y&=(\tan A)/(\tan A\tan B) - (\tan B)/(\tan A\tan B)\\\\y&=(\tan A-\tan B)/(\tan A\tan B)\\\\(1)/(y)&=(\tan A\tan B)/(\tan A-\tan B)\end{aligned}`

To find the value of cot(A - B), we can use the Tangent of Difference of Angles identity:

`\boxed{\begin{array}{l}\underline{\rm Tangent\;of\;Difference\; of \;Angles}\\\\\tan (A -B)=(\tan A -\tan B)/(1 +\tan A \tan B)\\\\\end{array}}`

`\textsf{As\;\;$\cot \theta=(1)/(\tan \theta)$,\;\;then $\cot(A-B)$ is:}`

`\cot(A -B)=(1 +\tan A \tan B)/(\tan A -\tan B)`

`\textsf{Apply the fraction rule} \quad (a+b)/(c)=(a)/(c)+(b)/(c):`

`\cot(A -B)=(1)/(\tan A -\tan B)+(\tan A \tan B)/(\tan A -\tan B)`

Substitute the given equation for x and the rewritten equation for 1/y:

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

Therefore, cot(A - B) is equal to:

`\large\boxed{(1)/(x)+(1)/(y)}`