If Sin ⁡ A = 9 41 \sin A = \frac{9}{41} Sin A = 41 9 ​ And Cos ⁡ B = 20 29 \cos B = \frac{20}{29} Cos B = 29 20 ​ , And Angles A And B Are In Quadrant I, Find The Value Of Tan ⁡ ( A + B \tan (A + B Tan ( A + B ].

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If sin⁡A=941\sin A = \frac{9}{41} and cos⁡B=2029\cos B = \frac{20}{29}, and angles A and B are in Quadrant I, find the value of tan⁡(A+B)\tan (A + B)

In this article, we will explore the problem of finding the value of tan⁡(A+B)\tan (A + B) given the values of sin⁡A\sin A and cos⁡B\cos B. We will use the given information to find the values of sin⁡B\sin B and cos⁡A\cos A, and then use the trigonometric identity for tan⁡(A+B)\tan (A + B) to find the final answer.

Before we begin, let's recall some important trigonometric identities that we will use in this problem.

  • sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1
  • tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}
  • tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

We are given that sin⁡A=941\sin A = \frac{9}{41} and cos⁡B=2029\cos B = \frac{20}{29}. We can use the Pythagorean identity to find the values of cos⁡A\cos A and sin⁡B\sin B.

  • cos⁡A=1−sin⁡2A=1−(941)2=4041\cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{9}{41}\right)^2} = \frac{40}{41}
  • sin⁡B=1−cos⁡2B=1−(2029)2=2129\sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{20}{29}\right)^2} = \frac{21}{29}

Now that we have the values of cos⁡A\cos A and sin⁡B\sin B, we can find the values of tan⁡A\tan A and tan⁡B\tan B.

  • tan⁡A=sin⁡Acos⁡A=9414041=940\tan A = \frac{\sin A}{\cos A} = \frac{\frac{9}{41}}{\frac{40}{41}} = \frac{9}{40}
  • tan⁡B=sin⁡Bcos⁡B=21292029=2120\tan B = \frac{\sin B}{\cos B} = \frac{\frac{21}{29}}{\frac{20}{29}} = \frac{21}{20}

Now that we have the values of tan⁡A\tan A and tan⁡B\tan B, we can use the trigonometric identity for tan⁡(A+B)\tan (A + B) to find the final answer.

  • tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B=940+21201−940⋅2120=940+84404040−1892000=93402000−1892000=934018112000=9340⋅20001811=93⋅501811=46501811\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{\frac{9}{40} + \frac{21}{20}}{1 - \frac{9}{40} \cdot \frac{21}{20}} = \frac{\frac{9}{40} + \frac{84}{40}}{\frac{40}{40} - \frac{189}{2000}} = \frac{\frac{93}{40}}{\frac{2000 - 189}{2000}} = \frac{\frac{93}{40}}{\frac{1811}{2000}} = \frac{93}{40} \cdot \frac{2000}{1811} = \frac{93 \cdot 50}{1811} = \frac{4650}{1811}

In this article, we used the given information to find the values of sin⁡B\sin B and cos⁡A\cos A, and then used the trigonometric identity for tan⁡(A+B)\tan (A + B) to find the final answer. We found that tan⁡(A+B)=46501811\tan (A + B) = \frac{4650}{1811}.

46501811\boxed{\frac{4650}{1811}}
Q&A: If sin⁡A=941\sin A = \frac{9}{41} and cos⁡B=2029\cos B = \frac{20}{29}, and angles A and B are in Quadrant I, find the value of tan⁡(A+B)\tan (A + B)

In our previous article, we explored the problem of finding the value of tan⁡(A+B)\tan (A + B) given the values of sin⁡A\sin A and cos⁡B\cos B. We used the given information to find the values of sin⁡B\sin B and cos⁡A\cos A, and then used the trigonometric identity for tan⁡(A+B)\tan (A + B) to find the final answer. In this article, we will answer some common questions related to this problem.

A: Angles A and B being in Quadrant I means that both angles are between 0 and 90 degrees. This is important because it affects the signs of the sine and cosine functions. In Quadrant I, both sine and cosine are positive.

A: We can use the Pythagorean identity to find the values of sin⁡B\sin B and cos⁡A\cos A. The Pythagorean identity states that sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. We can rearrange this equation to solve for sin⁡B\sin B and cos⁡A\cos A.

A: tan⁡A\tan A and tan⁡B\tan B are the ratios of the sine and cosine functions for angles A and B, respectively. tan⁡A=sin⁡Acos⁡A\tan A = \frac{\sin A}{\cos A} and tan⁡B=sin⁡Bcos⁡B\tan B = \frac{\sin B}{\cos B}. The values of tan⁡A\tan A and tan⁡B\tan B are used to find the value of tan⁡(A+B)\tan (A + B).

A: We can use the trigonometric identity for tan⁡(A+B)\tan (A + B) to find the final answer. The identity states that tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}. We can plug in the values of tan⁡A\tan A and tan⁡B\tan B to find the value of tan⁡(A+B)\tan (A + B).

A: The final answer for tan⁡(A+B)\tan (A + B) is 46501811\frac{4650}{1811}.

A: Yes, we can use this method to find the value of tan⁡(A−B)\tan (A - B). We can use the trigonometric identity for tan⁡(A−B)\tan (A - B), which states that tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}. We can plug in the values of tan⁡A\tan A and tan⁡B\tan B to find the value of tan⁡(A−B)\tan (A - B).

In this article, we answered some common questions related to the problem of finding the value of tan⁡(A+B)\tan (A + B) given the values of sin⁡A\sin A and cos⁡B\cos B. We used the given information to find the values of sin⁡B\sin B and cos⁡A\cos A, and then used the trigonometric identity for tan⁡(A+B)\tan (A + B) to find the final answer. We also discussed how to use this method to find the value of tan⁡(A−B)\tan (A - B).