Question

The value of \(\tan(\sin^{-1}(\frac{7}{25}))\) is equal to

• A. \(\frac{18}{25}\)
• B. \(\frac{24}{25}\)
• C. \(\frac{7}{24}\)
• D. \(\frac{3}{4}\)
• E. \(\frac{7}{18}\)

Answer 1

The Correct Option is C

We are given \( \sin^{-1} \left( \frac{7}{25} \right) \), which means we need to find the angle \( \theta \) such that \[ \sin \theta = \frac{7}{25}. \] We need to calculate \( \tan \theta \). Using the identity for sine and cosine in a right triangle, if \( \sin \theta = \frac{7}{25} \), then the opposite side is 7 and the hypotenuse is 25. We can find the adjacent side using the Pythagorean theorem: \[ \text{adjacent} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24. \] Thus, \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24}. \] Therefore, \[ \tan \left( \sin^{-1} \left( \frac{7}{25} \right) \right) = \frac{7}{24}. \]

The correct option is (C) : \(\frac{7}{24}\)

Answer 2

Let \(\theta = \sin^{-1}\left(\frac{7}{25}\right)\). This means \(\sin(\theta) = \frac{7}{25}\).

We want to find \(\tan(\theta)\).

We know that \(\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\). So, we can consider a right triangle where the opposite side is 7 and the hypotenuse is 25.

Let the adjacent side be \(a\). Using the Pythagorean theorem, we have \(a^2 + 7^2 = 25^2\), so \(a^2 + 49 = 625\), which means \(a^2 = 576\).

Taking the square root, we get \(a = \sqrt{576} = 24\). Since \(\theta = \sin^{-1}\left(\frac{7}{25}\right)\), \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\), which means \(\theta\) is in the first or fourth quadrant. Since \(\sin \theta > 0\), \(\theta\) is in the first quadrant, therefore \(a\) must be positive.

Then, \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24}\).

Therefore, \(\tan\left(\sin^{-1}\left(\frac{7}{25}\right)\right) = \frac{7}{24}\).