If tan A = 3/4, prove that Sin A Cos A = 12/25

Write Trig Identity: We know that tan A = sin A / cos A. Given tan A = 3/4, we can write sin A / cos A = 3/4.Apply Pythagorean Identity: To find sin A and cos A, we will use the Pythagorean identity: sin^2 A + cos^2 A = 1.Assume Values for sin A and cos A: Let's assume sin A = 3/k and cos A = 4/k for some k, because tan A = sin A / cos A = 3/4. We need to find the value of k.Substitute Values in Pythagorean Identity: Using the Pythagorean identity, we substitute sin A and cos A with the assumed values: (3/k)^2 + (4/k)^2 = 1.Solve for k: Solving the equation for k, we get 9/k^2 + 16/k^2 = 1.Find Value of k: Combining the terms, we have (9 + 16) / k^2 = 1, which simplifies to 25/k^2 = 1.Calculate sin A and cos A: Solving for k^2, we find k^2 = 25.Multiply sin A and cos A: Taking the square root of both sides, we get k = 5, since k is positive (k represents a length in the context of a right triangle).Prove Sin A Cos A: Now we can find sin A and cos A using the values of k: sin A = 3/k = 3/5 and cos A = 4/k = 4/5.To prove Sin A Cos A = 12/25, we multiply the values we found: sin A * cos A = (3/5) * (4/5).Calculating the product, we get sin A * cos A = 12/25, which is what we wanted to prove.

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If $\tan A = \frac{3}{4}$ , prove that $\sin A \cos A = \frac{12}{25}$

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Grade 6

Find a value using two-variable equations

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Q. If

\tan A = \frac{3}{4}

, prove that

\sin A \cos A = \frac{12}{25}

Write Trig Identity: We know that $\tan A = \frac{\sin A}{\cos A}$ . Given $\tan A = \frac{3}{4}$ , we can write $\frac{\sin A}{\cos A} = \frac{3}{4}$ .
Apply Pythagorean Identity: To find $\sin A$ and $\cos A$ , we will use the Pythagorean identity: $\sin^2 A + \cos^2 A = 1$ .
Assume Values for $\sin A$ and $\cos A$ : Let's assume $\sin A = \frac{3}{k}$ and $\cos A = \frac{4}{k}$ for some $k$ , because $\tan A = \frac{\sin A}{\cos A} = \frac{3}{4}$ . We need to find the value of $k$ .
Substitute Values in Pythagorean Identity: Using the Pythagorean identity, we substitute $\sin A$ and $\cos A$ with the assumed values: $(\frac{3}{k})^2 + (\frac{4}{k})^2 = 1$ .
Solve for k: Solving the equation for k, we get $\frac{9}{k^2} + \frac{16}{k^2} = 1$ .
Find Value of k: Combining the terms, we have $(9 + 16) / k^2 = 1$ , which simplifies to $25/k^2 = 1$ .
Calculate $\sin A$ and $\cos A$ : Solving for $k^2$ , we find $k^2 = 25$ .
Multiply $\sin A$ and $\cos A$ : Taking the square root of both sides, we get $k = 5$ , since $k$ is positive ( $k$ represents a length in the context of a right triangle).
Prove Sin A Cos A: Now we can find $\sin A$ and $\cos A$ using the values of $k$ : $\sin A = \frac{3}{k} = \frac{3}{5}$ and $\cos A = \frac{4}{k} = \frac{4}{5}$ .
Prove Sin A Cos A: Now we can find $\sin A$ and $\cos A$ using the values of $k$ : $\sin A = \frac{3}{k} = \frac{3}{5}$ and $\cos A = \frac{4}{k} = \frac{4}{5}$ .To prove $\sin A \cos A = \frac{12}{25}$ , we multiply the values we found: $\sin A \cdot \cos A = \left(\frac{3}{5}\right) \cdot \left(\frac{4}{5}\right)$ .
Prove Sin A Cos A: Now we can find $\sin A$ and $\cos A$ using the values of $k$ : $\sin A = \frac{3}{k} = \frac{3}{5}$ and $\cos A = \frac{4}{k} = \frac{4}{5}$ .To prove $\sin A \cos A = \frac{12}{25}$ , we multiply the values we found: $\sin A \cdot \cos A = \left(\frac{3}{5}\right) \cdot \left(\frac{4}{5}\right)$ .Calculating the product, we get $\sin A \cdot \cos A = \frac{12}{25}$ , which is what we wanted to prove.