Question

If cos A = 35/37 and sin B = 9/41 and angles A and B are in Quadrant I, find the valueof tan(A - B).

Answer 1

The given expression is tan(A - B)

Since tan (A - B) is equal to

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

Since the values of cos A and sin B are

`\begin{gathered} \cos A=(35)/(37) \\ \sin B=(9)/(41) \end{gathered}`

We will use the rules

`\begin{gathered} \tan ^2A=\sec ^2A-1\rightarrow(2) \\ \cot ^2B=\csc ^2B-1\rightarrow(3) \end{gathered}`

csc B = 1/sin B, cot B = 1/tan B, sec A = 1/cos A

`\begin{gathered} \csc B=(1)/((9)/(41))=(41)/(9) \\ \sec A=(1)/((35)/(37))=(37)/(35) \end{gathered}`

Substitute the value of csc B in rule (3) to find cot B

`\begin{gathered} \cot ^2B=((41)/(9))^2-1 \\ \cot ^2B=(1600)/(81) \\ \sqrt[]{\cot^2B}=\pm\sqrt[]{(1600)/(81)} \\ \cot B=(40)/(9) \end{gathered}`

Reciprocal it to find tan B

`\tan B=(1)/(\cot B)=(9)/(40)`

Substitute the value of sec A in rule (2) to find tan A

`\begin{gathered} \tan ^2A=((37)/(35))^2-1 \\ \tan ^2A=(144)/(1225) \\ \sqrt[]{\tan^2A}=\pm\sqrt[]{(144)/(1225)} \\ \tan A=(12)/(35) \end{gathered}`

Substitute the values of tan A and tan B in rule (1) above

`\begin{gathered} \tan (A-B)=((12)/(35)-(9)/(40))/(1+((12)/(35))((9)/(40))) \\ \tan (A-B)=(165)/(1508) \end{gathered}`

The value of tan(A - B) is 165/1508