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} $