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Asraf Ali

· started a discussion

· 1 Months ago

nice (Question: 111092.0 )

nice test series there is good combination of easy, moderate and tough level just there is little error in some answer key if it will take care then its like putting a cherry on the cake. good job!

Question:
If \(cosT=\cfrac{3}{5}\) and if \(sinR=\cfrac{8}{17},\) where T is in the fourth quadrant and R is in second quadrant, then cos (T-R) is equal to:
Options:
A) \(\cfrac{77}{85}\)
B) \(\cfrac{13}{85}\)
C) \(-\cfrac{13}{85}\)
D) \(-\cfrac{77}{85}\)
Solution:
Ans: (d)

We have

cos T = \(\cfrac{3}{5}\)

sin T=\(\sqrt{1-\cfrac{3^2}{5^2}}\)

= \(\sqrt{1-\cfrac{9}{25}}\)

=\(\sqrt{\cfrac{16}{25}}=-\cfrac{4}{5}\)              (since T is in IV quadrant)

sin R = \(\cfrac{8}{17}\)

cos R = \(\sqrt{1-\cfrac{8^2}{17^2}}\)

= \(\sqrt{1-\cfrac{64}{289}}\)

= \(\sqrt{\cfrac{225}{289}}=-\cfrac{15}{17}\)            (since R is in II quadrant)

Now, cos (T-R) =cos T cos R + sin T sin R

\(=\cfrac{3}{5}\times\cfrac{-15}{17}+\cfrac{-4}{5}\times\cfrac{8}{17}\)

= \(\left [ \begin{matrix}\cfrac{-45-32}{85}\end{matrix} \right ]\) =-\(\cfrac{77}{85}\)

Knowledge Expert

· commented

· 1 Months ago

Thank you,
Best Wishes.
Keep learning,
Team TR

Pranay SB

· commented

· 1 Months ago

Feeling the same bro
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