If tanθ = \(\frac{12}{5}\)

find the value of  \(\frac{24cotθ - 5cosθ+5}{13cosθ + 5tanθ-1}\) .

\(\frac{1}{8}\)  

tanθ = \(\frac{12}{5}\) = \(\frac{P}{B}\) (Triplet 5 , 12 , 13)

B = 13

Put than and find

⇒ \(\frac{24cotθ - 5cosθ+1}{13cosθ + 5tanθ-1}\)

⇒ \(\frac{24×\frac{5}{12} - 5 ×\frac{13}{5}+5}{13\frac{5}{13}+5×\frac{12}{5}-1}\) 

⇒ \(\frac{10 - 13 + 5 }{5 + 12 -1}\) 

⇒ \(\frac{2}{16}\)

= \(\frac{1}{8}\)