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Question: If acute angle A = 3B and $\sin A = \frac{4}{5}$ then $\frac{3 \sec B-4 \csc B}{2}$ is...

If acute angle A = 3B and sin⁡A=45\sin A = \frac{4}{5} then

3sec⁡B−4csc⁡B2\frac{3 \sec B-4 \csc B}{2} is

Answer

-5

Explanation

Solution

The problem asks us to evaluate the expression 3sec⁡B−4csc⁡B2\frac{3 \sec B-4 \csc B}{2}, given that A is an acute angle, A = 3B, and sin⁡A=45\sin A = \frac{4}{5}.

Step 1: Find cos⁡A\cos A. Given sin⁡A=45\sin A = \frac{4}{5}. Since A is an acute angle, we can use the identity sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1. cos⁡2A=1−sin⁡2A=1−(45)2=1−1625=25−1625=925\cos^2 A = 1 - \sin^2 A = 1 - \left(\frac{4}{5}\right)^2 = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25}. Since A is acute, cos⁡A\cos A must be positive. cos⁡A=925=35\cos A = \sqrt{\frac{9}{25}} = \frac{3}{5}.

Step 2: Use the relation A = 3B. We have sin⁡A=sin⁡(3B)=45\sin A = \sin(3B) = \frac{4}{5} and cos⁡A=cos⁡(3B)=35\cos A = \cos(3B) = \frac{3}{5}. Consider the identity for sin⁡(A−B)\sin(A-B): sin⁡(A−B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(A-B) = \sin A \cos B - \cos A \sin B. Substitute A=3BA = 3B: sin⁡(3B−B)=sin⁡(2B)\sin(3B-B) = \sin(2B). So, sin⁡(2B)=sin⁡Acos⁡B−cos⁡Asin⁡B\sin(2B) = \sin A \cos B - \cos A \sin B. Substitute the values of sin⁡A\sin A and cos⁡A\cos A: sin⁡(2B)=45cos⁡B−35sin⁡B\sin(2B) = \frac{4}{5} \cos B - \frac{3}{5} \sin B.

Step 3: Use the double angle identity for sin⁡(2B)\sin(2B). We know sin⁡(2B)=2sin⁡Bcos⁡B\sin(2B) = 2 \sin B \cos B. Equating the two expressions for sin⁡(2B)\sin(2B): 2sin⁡Bcos⁡B=45cos⁡B−35sin⁡B2 \sin B \cos B = \frac{4}{5} \cos B - \frac{3}{5} \sin B.

Step 4: Rearrange the equation to find a relationship between sin⁡B\sin B and cos⁡B\cos B. Multiply the entire equation by 5 to eliminate the denominators: 10sin⁡Bcos⁡B=4cos⁡B−3sin⁡B10 \sin B \cos B = 4 \cos B - 3 \sin B.

Step 5: Evaluate the given expression. The expression to evaluate is 3sec⁡B−4csc⁡B2\frac{3 \sec B-4 \csc B}{2}. Rewrite sec⁡B\sec B as 1cos⁡B\frac{1}{\cos B} and csc⁡B\csc B as 1sin⁡B\frac{1}{\sin B}: 12(3cos⁡B−4sin⁡B)\frac{1}{2} \left( \frac{3}{\cos B} - \frac{4}{\sin B} \right). Combine the terms inside the parenthesis by finding a common denominator: 12(3sin⁡B−4cos⁡Bsin⁡Bcos⁡B)\frac{1}{2} \left( \frac{3 \sin B - 4 \cos B}{\sin B \cos B} \right).

From Step 4, we have 10sin⁡Bcos⁡B=4cos⁡B−3sin⁡B10 \sin B \cos B = 4 \cos B - 3 \sin B. Notice that the numerator in the expression we want to evaluate is 3sin⁡B−4cos⁡B3 \sin B - 4 \cos B, which is the negative of the right side of the equation from Step 4. So, 3sin⁡B−4cos⁡B=−(4cos⁡B−3sin⁡B)=−10sin⁡Bcos⁡B3 \sin B - 4 \cos B = -(4 \cos B - 3 \sin B) = -10 \sin B \cos B.

Substitute this back into the expression: 12(−10sin⁡Bcos⁡Bsin⁡Bcos⁡B)\frac{1}{2} \left( \frac{-10 \sin B \cos B}{\sin B \cos B} \right).

Since A is an acute angle, A∈(0,90∘)A \in (0, 90^\circ). Since A=3BA = 3B, 3B∈(0,90∘)3B \in (0, 90^\circ), which implies B∈(0,30∘)B \in (0, 30^\circ). For B∈(0,30∘)B \in (0, 30^\circ), sin⁡B≠0\sin B \neq 0 and cos⁡B≠0\cos B \neq 0. Therefore, sin⁡Bcos⁡B≠0\sin B \cos B \neq 0. We can cancel sin⁡Bcos⁡B\sin B \cos B from the numerator and denominator: 12(−10)=−5\frac{1}{2} (-10) = -5.