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If cos A=(4)/(5) and tan B=(11)/(60) and angles A and B are in Quadrant I, find the value of tan(A+B). Answer:

If cos⁡A=45 \cos A=\frac{4}{5} and tan⁡B=1160 \tan B=\frac{11}{60} and angles A and B are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) .\newlineAnswer:

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Q. If cos⁡A=45 \cos A=\frac{4}{5} and tan⁡B=1160 \tan B=\frac{11}{60} and angles A and B are in Quadrant I, find the value of tan⁡(A+B) \tan (A+B) .\newlineAnswer:
  1. Find sin⁡A\sin A: We know that cos⁡A=45\cos A = \frac{4}{5}. To find tan⁡A\tan A, we need to find sin⁡A\sin A using the Pythagorean identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1.\newlineSubstitute cos⁡A=45\cos A = \frac{4}{5} into the identity:\newline(\sin A)^\(2 + \left(\frac{44}{55}\right)^22 = 11
  2. Simplify equation: Simplify the equation to find (sin⁡A)2(\sin A)^2:(sin⁡A)2+1625=1(\sin A)^2 + \frac{16}{25} = 1(sin⁡A)2=1−1625(\sin A)^2 = 1 - \frac{16}{25}(sin⁡A)2=2525−1625(\sin A)^2 = \frac{25}{25} - \frac{16}{25}(sin⁡A)2=925(\sin A)^2 = \frac{9}{25}
  3. Take square root: Take the square root of both sides to find sin⁡A\sin A:sin⁡A=±925\sin A = \pm\sqrt{\frac{9}{25}}Since angle A is in Quadrant I, where sine is positive, we choose the positive root:sin⁡A=35\sin A = \frac{3}{5}
  4. Find tan⁡A\tan A: Now we can find tan⁡A\tan A using sin⁡A\sin A and cos⁡A\cos A:
    tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A}
    tan⁡A=3/54/5\tan A = \frac{3/5}{4/5}
    tan⁡A=34\tan A = \frac{3}{4}
  5. Use angle sum identity: We are given tan⁡B=1160\tan B = \frac{11}{60}. To find tan⁡(A+B)\tan(A+B), we use the angle sum identity for tangent:\newlinetan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡A⋅tan⁡B\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}\newlineSubstitute tan⁡A=34\tan A = \frac{3}{4} and tan⁡B=1160\tan B = \frac{11}{60} into the identity:\newlinetan⁡(A+B)=34+11601−(34)⋅(1160)\tan(A+B) = \frac{\frac{3}{4} + \frac{11}{60}}{1 - \left(\frac{3}{4}\right) \cdot \left(\frac{11}{60}\right)}
  6. Find common denominator: Find a common denominator and simplify the numerator and denominator:\newlinetan⁡(A+B)=45/60+11/601−33/240\tan(A+B) = \frac{45/60 + 11/60}{1 - 33/240}\newlinetan⁡(A+B)=56/601−33/240\tan(A+B) = \frac{56/60}{1 - 33/240}\newlinetan⁡(A+B)=56/60(240/240)−(33/240)\tan(A+B) = \frac{56/60}{(240/240) - (33/240)}\newlinetan⁡(A+B)=56/60207/240\tan(A+B) = \frac{56/60}{207/240}
  7. Simplify fractions: Simplify the fractions and perform the division:\newlinetan⁡(A+B)=1415/207240\tan(A+B) = \frac{14}{15} / \frac{207}{240}\newlinetan⁡(A+B)=1415×240207\tan(A+B) = \frac{14}{15} \times \frac{240}{207}\newlinetan⁡(A+B)=14×24015×207\tan(A+B) = \frac{14 \times 240}{15 \times 207}
  8. Perform division: Simplify the multiplication and division: tan⁡(A+B)=33603105\tan(A+B) = \frac{3360}{3105} Now we simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 1515: tan⁡(A+B)=224207\tan(A+B) = \frac{224}{207}

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