Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

If 
sin A=(9)/(41) and 
cos B=(21)/(29) and angles A and B are in Quadrant I, find the value of 
tan(A-B).
Answer:

If sinA=941 \sin A=\frac{9}{41} and cosB=2129 \cos B=\frac{21}{29} and angles A and B are in Quadrant I, find the value of tan(AB) \tan (A-B) .\newlineAnswer:

Full solution

Q. If sinA=941 \sin A=\frac{9}{41} and cosB=2129 \cos B=\frac{21}{29} and angles A and B are in Quadrant I, find the value of tan(AB) \tan (A-B) .\newlineAnswer:
  1. Find cosA\cos A: Use the Pythagorean identity to find cosA\cos A.cosA=1sin2A\cos A = \sqrt{1 - \sin^2 A}cosA=1(941)2\cos A = \sqrt{1 - (\frac{9}{41})^2}cosA=1811681\cos A = \sqrt{1 - \frac{81}{1681}}cosA=16001681\cos A = \sqrt{\frac{1600}{1681}}cosA=4041\cos A = \frac{40}{41}
  2. Find sinB\sin B: Use the Pythagorean identity to find sinB\sin B.
    sinB=1cos2B\sin B = \sqrt{1 - \cos^2 B}
    sinB=1(2129)2\sin B = \sqrt{1 - \left(\frac{21}{29}\right)^2}
    sinB=1441841\sin B = \sqrt{1 - \frac{441}{841}}
    sinB=400841\sin B = \sqrt{\frac{400}{841}}
    sinB=2029\sin B = \frac{20}{29}
  3. Find tan(AB)\tan(A-B): Use the angle difference identity for tangent to find tan(AB)\tan(A-B).tan(AB)=tanAtanB1+tanAtanB\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}tanA=sinAcosA=9/4140/41=940\tan A = \frac{\sin A}{\cos A} = \frac{9/41}{40/41} = \frac{9}{40}tanB=sinBcosB=20/2921/29=2021\tan B = \frac{\sin B}{\cos B} = \frac{20/29}{21/29} = \frac{20}{21}tan(AB)=(940)(2021)1+(940)(2021)\tan(A-B) = \frac{(\frac{9}{40}) - (\frac{20}{21})}{1 + (\frac{9}{40}) \cdot (\frac{20}{21})}
  4. Simplify tan(AB)\tan(A-B): Simplify the expression for tan(AB)\tan(A-B).
    tan(AB)=(940)(2021)1+(940)(2021)\tan(A-B) = \frac{(\frac{9}{40}) - (\frac{20}{21})}{1 + (\frac{9}{40}) \cdot (\frac{20}{21})}
    tan(AB)=(921)(2040)4021/(1+9204021)\tan(A-B) = \frac{(9\cdot21) - (20\cdot40)}{40\cdot21} / (1 + \frac{9\cdot20}{40\cdot21})
    tan(AB)=189800840/(1+180840)\tan(A-B) = \frac{189 - 800}{840} / \left(1 + \frac{180}{840}\right)
    tan(AB)=611840/(1+180840)\tan(A-B) = \frac{-611}{840} / \left(1 + \frac{180}{840}\right)
    tan(AB)=611840/(840+180840)\tan(A-B) = \frac{-611}{840} / \left(\frac{840 + 180}{840}\right)
    tan(AB)=611840/(1020840)\tan(A-B) = \frac{-611}{840} / \left(\frac{1020}{840}\right)
    tan(AB)=6118408401020\tan(A-B) = \frac{-611}{840} \cdot \frac{840}{1020}
    tan(AB)=6111020\tan(A-B) = \frac{-611}{1020}
    tan(AB)=6111020\tan(A-B) = \frac{-611}{1020}
  5. Simplify fraction: Simplify the fraction 611/1020-611 / 1020. Both the numerator and the denominator can be divided by 6161. tan(AB)=611/1020\tan(A-B) = -611 / 1020 tan(AB)=10/17\tan(A-B) = -10 / 17

More problems from Find trigonometric ratios using multiple identities