If $\sin A=\frac{12}{37}$ and $\tan B=\frac{7}{24}$ and angles A and B are in Quadrant I, find the value of $\tan (A-B)$ . $\newline$ Answer:

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Find trigonometric ratios using multiple identities

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Q. If

\sin A=\frac{12}{37}

and

\tan B=\frac{7}{24}

and angles A and B are in Quadrant I, find the value of

\tan (A-B)

\newline

Answer:

Use Pythagorean Identity: We know that $\sin A = \frac{12}{37}$ . To find $\cos A$ , we use the Pythagorean identity $\sin^2 A + \cos^2 A = 1$ . $\newline$ Substitute $\sin A$ into the identity: $\newline$ $(\frac{12}{37})^2 + \cos^2 A = 1$ .
Calculate $\cos A$ : Calculate $(12/37)^2$ and simplify the equation to find $\cos^2 A$ : $\$12/37)^2 = 144/1369$ . $\cos^2 A = 1 - 144/1369$ .
Find $\cos B$ : Subtract $\frac{144}{1369}$ from $1$ to find $\cos^2 A$ :
$\cos^2 A = \frac{1369}{1369} - \frac{144}{1369}$ .
$\cos^2 A = \frac{1225}{1369}$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{1225/1369}$ . $\newline$ $\cos A = 35/37$ .We know that $\tan B = 7/24$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .Now we have $\cos A = \frac{35}{37}$ , $A$ $6$ , $A$ $4$ , and $\tan B = \frac{7}{24}$ . We can use the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ Since we have $\cos A$ $1$ and $\cos A$ , we can find $\cos A$ $3$ : $\newline$ $\cos A$ $4$ . $\newline$ $\cos A$ $5$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .Now we have $\cos A = \frac{35}{37}$ , $A$ $6$ , $A$ $4$ , and $\tan B = \frac{7}{24}$ . We can use the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ Since we have $\cos A$ $1$ and $\cos A$ , we can find $\cos A$ $3$ : $\newline$ $\cos A$ $4$ . $\newline$ $\cos A$ $5$ .Calculate $\cos A$ $3$ by dividing $\cos A$ $1$ by $\cos A$ : $\newline$ $\cos A$ $9$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $0$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .Now we have $\cos A = \frac{35}{37}$ , $A$ $6$ , $A$ $4$ , and $\tan B = \frac{7}{24}$ . We can use the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ Since we have $\cos A$ $1$ and $\cos A$ , we can find $\cos A$ $3$ : $\newline$ $\cos A$ $4$ . $\newline$ $\cos A$ $5$ .Calculate $\cos A$ $3$ by dividing $\cos A$ $1$ by $\cos A$ : $\newline$ $\cos A$ $9$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $0$ .Substitute $\cos A$ $3$ and $\cos A = \sqrt{\frac{1225}{1369}}$ $2$ into the formula for $\tan(A-B)$ : $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $4$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .Now we have $\cos A = \frac{35}{37}$ , $A$ $6$ , $A$ $4$ , and $\tan B = \frac{7}{24}$ . We can use the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ Since we have $\cos A$ $1$ and $\cos A$ , we can find $\cos A$ $3$ : $\newline$ $\cos A$ $4$ . $\newline$ $\cos A$ $5$ .Calculate $\cos A$ $3$ by dividing $\cos A$ $1$ by $\cos A$ : $\newline$ $\cos A$ $9$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $0$ .Substitute $\cos A$ $3$ and $\cos A = \sqrt{\frac{1225}{1369}}$ $2$ into the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $5$ .Calculate the numerator and denominator separately: $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $6$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $7$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $8$ . $\newline$ Denominator: $\cos A = \sqrt{\frac{1225}{1369}}$ $9$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $0$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $1$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .Now we have $\cos A = \frac{35}{37}$ , $A$ $6$ , $A$ $4$ , and $\tan B = \frac{7}{24}$ . We can use the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ Since we have $\cos A$ $1$ and $\cos A$ , we can find $\cos A$ $3$ : $\newline$ $\cos A$ $4$ . $\newline$ $\cos A$ $5$ .Calculate $\cos A$ $3$ by dividing $\cos A$ $1$ by $\cos A$ : $\newline$ $\cos A$ $9$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $0$ .Substitute $\cos A$ $3$ and $\cos A = \sqrt{\frac{1225}{1369}}$ $2$ into the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $5$ .Calculate the numerator and denominator separately: $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $6$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $7$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $8$ . $\newline$ Denominator: $\cos A = \sqrt{\frac{1225}{1369}}$ $9$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $0$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $1$ .Now, divide the numerator by the denominator to find $\tan(A-B)$ : $\newline$ $\cos A = \frac{35}{37}$ $3$ . $\newline$ $\cos A = \frac{35}{37}$ $4$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .Now we have $\cos A = \frac{35}{37}$ , $A$ $6$ , $A$ $4$ , and $\tan B = \frac{7}{24}$ . We can use the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ Since we have $\cos A$ $1$ and $\cos A$ , we can find $\cos A$ $3$ : $\newline$ $\cos A$ $4$ . $\newline$ $\cos A$ $5$ .Calculate $\cos A$ $3$ by dividing $\cos A$ $1$ by $\cos A$ : $\newline$ $\cos A$ $9$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $0$ .Substitute $\cos A$ $3$ and $\cos A = \sqrt{\frac{1225}{1369}}$ $2$ into the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $5$ .Calculate the numerator and denominator separately: $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $6$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $7$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $8$ . $\newline$ Denominator: $\cos A = \sqrt{\frac{1225}{1369}}$ $9$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $0$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $1$ .Now, divide the numerator by the denominator to find $\tan(A-B)$ : $\newline$ $\cos A = \frac{35}{37}$ $3$ . $\newline$ $\cos A = \frac{35}{37}$ $4$ .Simplify the expression by canceling out the common factor of $\cos A = \frac{35}{37}$ $5$ : $\newline$ $\cos A = \frac{35}{37}$ $6$ . $\newline$ $\cos A = \frac{35}{37}$ $7$ .
Calculate $\tan(A-B)$ : Take the square root of $\cos^2 A$ to find $\cos A$ . Since $A$ is in Quadrant I, $\cos A$ is positive: $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ . $\newline$ $\cos A = \frac{35}{37}$ .We know that $\tan B = \frac{7}{24}$ . To find $\cos B$ , we use the identity $1 + \tan^2 B = \sec^2 B$ . $\newline$ First, find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $1$ .Calculate $\cos^2 A$ $2$ and add it to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos^2 A$ $5$ . $\newline$ $\cos^2 A$ $6$ .Add $\cos^2 A$ $7$ to $\cos^2 A$ $3$ to find $\cos^2 A$ $0$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A$ $1$ .Take the square root of $\cos^2 A$ $0$ to find $\cos A$ $3$ . Since $\cos A$ $4$ is in Quadrant I, $\cos A$ $3$ is positive: $\newline$ $\cos A$ $6$ . $\newline$ $\cos A$ $7$ .To find $\cos B$ , we use the reciprocal of $\cos A$ $3$ : $\newline$ $A$ $0$ . $\newline$ $A$ $1$ .Calculate the reciprocal of $\cos A$ $3$ to find $\cos B$ : $\newline$ $A$ $4$ .Now we have $\cos A = \frac{35}{37}$ , $A$ $6$ , $A$ $4$ , and $\tan B = \frac{7}{24}$ . We can use the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ Since we have $\cos A$ $1$ and $\cos A$ , we can find $\cos A$ $3$ : $\newline$ $\cos A$ $4$ . $\newline$ $\cos A$ $5$ .Calculate $\cos A$ $3$ by dividing $\cos A$ $1$ by $\cos A$ : $\newline$ $\cos A$ $9$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $0$ .Substitute $\cos A$ $3$ and $\cos A = \sqrt{\frac{1225}{1369}}$ $2$ into the formula for $\tan(A-B)$ : $\newline$ $\cos A$ $0$ . $\newline$ $\cos A = \sqrt{\frac{1225}{1369}}$ $5$ .Calculate the numerator and denominator separately: $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $6$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $7$ . $\newline$ Numerator: $\cos A = \sqrt{\frac{1225}{1369}}$ $8$ . $\newline$ Denominator: $\cos A = \sqrt{\frac{1225}{1369}}$ $9$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $0$ . $\newline$ Denominator: $\cos A = \frac{35}{37}$ $1$ .Now, divide the numerator by the denominator to find $\tan(A-B)$ : $\newline$ $\cos A = \frac{35}{37}$ $3$ . $\newline$ $\cos A = \frac{35}{37}$ $4$ .Simplify the expression by canceling out the common factor of $\cos A = \frac{35}{37}$ $5$ : $\newline$ $\cos A = \frac{35}{37}$ $6$ . $\newline$ $\cos A = \frac{35}{37}$ $7$ .Simplify the fraction to get the final answer: $\newline$ $\cos A = \frac{35}{37}$ $7$ .

If sin⁡A=1237 \sin A=\frac{12}{37} sinA=3712​ and tan⁡B=724 \tan B=\frac{7}{24} tanB=247​ and angles A and B are in Quadrant I, find the value of tan⁡(A−B) \tan (A-B) tan(A−B).\newlineAnswer:

If $\sin A=\frac{12}{37}$ and $\tan B=\frac{7}{24}$ and angles A and B are in Quadrant I, find the value of $\tan (A-B)$ . $\newline$ Answer: