Question

Given: sine (a) = startfraction 5 over 13 endfraction, startfraction pi over 2 endfraction < a < pi and tangent (b) = negative startroot 13 endroot, startfraction pi over 2 endfraction < b < pi what is tan(a – b)? startfraction 5 12 startroot 13 endroot over 12 minus 5 startroot 13 endroot endfraction startfraction 12 minus 5 startroot 13 endroot over 5 12 startroot 13 endroot endfraction startfraction 12 5 startroot 13 endroot over negative 5 12 startroot 13 endroot endfraction startfraction negative 5 12 startroot 13 endroot over 12 5 startroot 13 endroot endfraction

Answer 1

To solve for tan(a - b), we use trigonometric identities to find the missing cos(a) and sin(b). We then apply the difference identity for tangent with the found values. The correct answer involves rationalizing the ratio using these identities and values.

Step-by-step explanation:

To find tan(a − b), we can use the identity for the tangent of the difference of two angles:

tan(a − b) = \(\frac{tan a - tan b}{1 + tan a \cdot tan b}\)

Since we are given sin(a) = \(\frac{5}{13}) and tangent (b) = -\(\sqrt{13}), we need to find cos(a) and sin(b) to use the given identity.

For angle a, located in the second quadrant (where sin is positive and cos is negative), we use the Pythagorean identity cos2(a) = 1 − sin2(a) to find cos(a). For angle b, also in the second quadrant (where tan is negative), we can construct a right triangle with sides such that the opposite side (y) is -\(\sqrt{13}) and the adjacent side (x) is -1 (to have a negative tangent of \(\sqrt{13})), with a hypotenuse (h) of 2, using the Pythagorean theorem. This gives us sin(b) = -\(\sqrt{13})/2.

Now, we calculate:

cos(a) = -\(\sqrt{1 - \(\frac{5}{13})2}) = -\(\sqrt{\(\frac{169 - 25}{169}}) = -\(\frac{12}{13})

sin(b) = y/h = -\(\sqrt{13})/2

Using the identity:

tan(a − b) = \(\frac{-\(\frac{5}{13}) - (-\(\sqrt{13})}{1 + (-\(\frac{5}{13}) \cdot (-\(\sqrt{13}))}

The correct answer is \(\frac{-5 \(\sqrt{13}) / 12}{12 + 5 \(\sqrt{13})}