If cos A = –12/13 and cot B = 24/7, where A lies in the second quadrant and B in the third quadrant, find the values of the following: ​

Question

If cos A = –\(\frac{12}{13}\) and cot B = \(\frac{24}{7}\), where A lies in the second quadrant and B in the third quadrant, find the values of the following: 

(i) sin(A +B) 

(ii) cos(A +B) 

(iii) tan(A +B)

Answer 1

Given cos A = -\(\frac{12}{13}\) And cotB = \(\frac{24}{7}\)

A lies in second quadrant And B in the third quadrant. 

The sine function is positive in the second quadrant and in the third quadrant,Both sine And cosine functions are negative. 

We know that

sin A = \(\sqrt{1-cos^2A}\) and sin B = -\(\frac1{\sqrt{1+cot^2B}}\)

⇒ cos B = -\(\sqrt{1-sin^2B}\) = -\(\sqrt{1-(-\frac{7}{25})^2}\)

= - \(\sqrt{1-\frac{49}{625}}\) = - \(\sqrt{\frac{576}{625}}\)

= - \(\frac{24}{25}\)

Now, 

(i) Sin(A +B) 

We know that sin(A +B) = sinA cosB + cosA sinB

(ii) Cos(A +B) 

We know that cos(A +B) = cosA cosB - sinA sinB

(iii) Tan(A +B) 

We know that tan(A + B) = \(\frac{sin(A+B)}{cos(A+B)}\)