Question

If \[ \tan\alpha=-\frac{7}{24}, \qquad \sec\beta=\frac{61}{60} \] and both \(\alpha,\beta\) lie in the same quadrant, then \[ \cos(\alpha+\beta)= \]

• A. \(\dfrac{1363}{1525}\)
• B. \(\dfrac{1236}{1525}\)
• C. \(\dfrac{23}{24}\)
• D. \(\dfrac{55}{61}\)

Hint:

When trigonometric ratios and quadrant information are given, first determine the signs of sine and cosine. Then use standard angle-sum identities directly.

Answer 1

The Correct Option is A

Concept: Use the identities \[ \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta. \] First determine \(\sin\alpha,\cos\alpha,\sin\beta,\cos\beta\) from the given values and the quadrant information.

Step 1: Find \(\sin\alpha\) and \(\cos\alpha\). Given \[ \tan\alpha=-\frac{7}{24}. \] Using the Pythagorean triple \[ 7^2+24^2=25^2, \] we get \[ |\sin\alpha|=\frac{7}{25}, \qquad |\cos\alpha|=\frac{24}{25}. \] Since \(\tan\alpha\lt 0\), \(\alpha\) lies either in Quadrant II or IV. Also, \[ \sec\beta=\frac{61}{60}\gt 0, \] so \(\beta\) lies either in Quadrant I or IV. As \(\alpha\) and \(\beta\) are in the same quadrant, both must lie in Quadrant IV. Therefore, \[ \sin\alpha=-\frac{7}{25}, \qquad \cos\alpha=\frac{24}{25}. \]

Step 2: Find \(\sin\beta\) and \(\cos\beta\). Given \[ \sec\beta=\frac{61}{60}, \] hence \[ \cos\beta=\frac{60}{61}. \] Using \[ \sin^2\beta+\cos^2\beta=1, \] \[ \sin\beta = \pm\sqrt{1-\left(\frac{60}{61}\right)^2} = \pm\frac{11}{61}. \] Since \(\beta\) is in Quadrant IV, \[ \sin\beta=-\frac{11}{61}. \]

Step 3: Calculate \(\cos(\alpha+\beta)\). \[ \cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta. \] \[ = \left(\frac{24}{25}\right)\left(\frac{60}{61}\right) - \left(-\frac{7}{25}\right)\left(-\frac{11}{61}\right). \] \[ = \frac{1440}{1525} - \frac{77}{1525}. \] \[ = \frac{1363}{1525}. \] Therefore, \[ \boxed{\cos(\alpha+\beta)=\frac{1363}{1525}} \] \[ \boxed{\text{Answer = (A)}} \]