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Question: If $\frac{sin A}{sin B} = \frac{\sqrt{3}}{2}$ and $\frac{cos A}{cos B} = \frac{\sqrt{5}}{2}, 0 < A, ...

If sinAsinB=32\frac{sin A}{sin B} = \frac{\sqrt{3}}{2} and cosAcosB=52,0<A,B<π2\frac{cos A}{cos B} = \frac{\sqrt{5}}{2}, 0 < A, B < \frac{\pi}{2} then tan A + tan B is equal to:

A

35\sqrt{\frac{3}{5}}

B

53\sqrt{\frac{5}{3}}

C

3+55\frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}

D

3+53\frac{\sqrt{3} + \sqrt{5}}{\sqrt{3}}

Answer

(c)

Explanation

Solution

The problem asks us to find the value of tan⁡A+tan⁡B\tan A + \tan B given two trigonometric ratios and the range for angles A and B.

Given:

  1. sin⁡Asin⁡B=32\frac{\sin A}{\sin B} = \frac{\sqrt{3}}{2}
  2. cos⁡Acos⁡B=52\frac{\cos A}{\cos B} = \frac{\sqrt{5}}{2}
  3. 0<A,B<π20 < A, B < \frac{\pi}{2}

From equation (1), we can write sin⁡A=32sin⁡B\sin A = \frac{\sqrt{3}}{2} \sin B. From equation (2), we can write cos⁡A=52cos⁡B\cos A = \frac{\sqrt{5}}{2} \cos B.

We know the fundamental trigonometric identity: sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. Substitute the expressions for sin⁡A\sin A and cos⁡A\cos A in terms of sin⁡B\sin B and cos⁡B\cos B into this identity: (32sin⁡B)2+(52cos⁡B)2=1(\frac{\sqrt{3}}{2} \sin B)^2 + (\frac{\sqrt{5}}{2} \cos B)^2 = 1 34sin⁡2B+54cos⁡2B=1\frac{3}{4} \sin^2 B + \frac{5}{4} \cos^2 B = 1

Multiply the entire equation by 4 to clear the denominators: 3sin⁡2B+5cos⁡2B=43 \sin^2 B + 5 \cos^2 B = 4

Now, we can express this equation in terms of tan⁡B\tan B. Divide the entire equation by cos⁡2B\cos^2 B. Since 0<B<π20 < B < \frac{\pi}{2}, cos⁡B≠0\cos B \neq 0. 3sin⁡2Bcos⁡2B+5cos⁡2Bcos⁡2B=4cos⁡2B\frac{3 \sin^2 B}{\cos^2 B} + \frac{5 \cos^2 B}{\cos^2 B} = \frac{4}{\cos^2 B} 3tan⁡2B+5=4sec⁡2B3 \tan^2 B + 5 = 4 \sec^2 B

Using the identity sec⁡2B=1+tan⁡2B\sec^2 B = 1 + \tan^2 B: 3tan⁡2B+5=4(1+tan⁡2B)3 \tan^2 B + 5 = 4 (1 + \tan^2 B) 3tan⁡2B+5=4+4tan⁡2B3 \tan^2 B + 5 = 4 + 4 \tan^2 B

Rearrange the terms to solve for tan⁡2B\tan^2 B: 5−4=4tan⁡2B−3tan⁡2B5 - 4 = 4 \tan^2 B - 3 \tan^2 B 1=tan⁡2B1 = \tan^2 B

Since 0<B<π20 < B < \frac{\pi}{2}, tan⁡B\tan B must be positive. Therefore, tan⁡B=1\tan B = 1

Next, we need to find tan⁡A\tan A. We can express tan⁡A\tan A in terms of tan⁡B\tan B: tan⁡A=sin⁡Acos⁡A=32sin⁡B52cos⁡B=35sin⁡Bcos⁡B=35tan⁡B\tan A = \frac{\sin A}{\cos A} = \frac{\frac{\sqrt{3}}{2} \sin B}{\frac{\sqrt{5}}{2} \cos B} = \frac{\sqrt{3}}{\sqrt{5}} \frac{\sin B}{\cos B} = \frac{\sqrt{3}}{\sqrt{5}} \tan B

Substitute the value of tan⁡B=1\tan B = 1: tan⁡A=35×1=35\tan A = \frac{\sqrt{3}}{\sqrt{5}} \times 1 = \frac{\sqrt{3}}{\sqrt{5}}

Finally, we need to calculate tan⁡A+tan⁡B\tan A + \tan B: tan⁡A+tan⁡B=35+1\tan A + \tan B = \frac{\sqrt{3}}{\sqrt{5}} + 1 To combine these terms, find a common denominator: tan⁡A+tan⁡B=35+55=3+55\tan A + \tan B = \frac{\sqrt{3}}{\sqrt{5}} + \frac{\sqrt{5}}{\sqrt{5}} = \frac{\sqrt{3} + \sqrt{5}}{\sqrt{5}}

Comparing this result with the given options, the calculated value matches option (c).