Question

If $\cos T=\frac{3}{5}$ and if $\sin R=\frac{8}{17}$, where T is in the fourth quadrant and R is in the second quadrant, then cos (T - R) is equal to:

• A. $\frac{77}{85}$
• B. $\frac{13}{85}$
• C. $-\frac{13}{85}$
• D. $-\frac{77}{85}$

Answer 1

Answer: (D)

As given, cos T = 3/5
${\sin }^{2}T=1-\frac{3^2}{5^2}=1-\frac{9}{25}$
So, $\sin T=\frac{16}{25}=-\frac{4}{5}$
[since T is in IV quad. + ve value is ignored.]
Also, given, sin R = $\frac{8}{17}$
$\Rightarrow {\cos }^2=1-\frac{8^2}{17^2}=1-\frac{64}{289}=\frac{225}{289}$
So, $\cos R=-\frac{15}{17}$
[Since R is in II quad. + ve value is ignored]
Now, cos (T - R) = cos T cos R + sin T sin R
$=\frac{3}{5}\times \frac{-15}{17}-\frac{4}{5}\times \frac{8}{17}=-\frac{45}{85}-\frac{32}{85}$
$=-\left[\frac{45+32}{85}\right]=-\frac{77}{85}$