If $tan A + sec A = \frac{3}{2}$ and A is an acute, then the value of $\frac{10cotA +13cosA}{12ranA+5cosecA}$ is :

2

tan A + secA = \(\frac{3}{2}\)    -----(1)

( sec²A - tan²A = 1 So, secA - tanA = \(\frac{1}{secA + tanA}\) )

So, secA - tanA = \(\frac{2}{3}\)        -----(2)

On adding equation 1 and 2.

2 secA = \(\frac{3}{2}\) + \(\frac{2}{3}\) 

secA = \(\frac{13}{12}\) 

{ secA = \(\frac{H}{B}\)  }

By using pyhtagoras theorem,

P² + B² = H²

P² + 12² = 13²

P = 5

Now,

\(\frac{10cotA + 13cosA }{12tanA + 5 cosecA }\) 

= \(\frac{10 × \(\frac{12}{5}\) + 13 × \(\frac{12}{13}\) }{12× \(\frac{5}{12}\) + 5 × \(\frac{13}{5}\)}\)  

= \(\frac{24 + 12 }{5 + 13}\) 

= \(\frac{36 }{18}\) 

= 2