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

$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}\rightarrow(1)$

Since the values of cos A and sin B are

$\begin{gathered} \cos A=\frac{35}{37} \\ \sin B=\frac{9}{41} \end{gathered}$

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

$\begin{gathered} \csc B=\frac{1}{\frac{9}{41}}=\frac{41}{9} \\ \sec A=\frac{1}{\frac{35}{37}}=\frac{37}{35} \end{gathered}$

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

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

Reciprocal it to find tan B

$\tan B=\frac{1}{\cot B}=\frac{9}{40}$

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

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

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

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

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