If $tan A + sec A = \frac{3}{2}$ and A is an acute, then the value of $\frac{10cotA +13cosA}{12ranA+5cosecA}$ is :
Answer & explanation
Correct answer: option 4
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