For $$\theta$$ being an acute angle, if $$\cosec \theta = 1.25$$, then the value of $$\frac{4 \tan \theta - 5 \cos \theta}{\sec \theta + 4 \cot \theta}$$ is equal to:

we have cosec$$\theta\ =\frac{5}{4}$$
so we get $$\sin\theta\ =\frac{4}{5}$$

we know, $$\cos\ \theta\ =\sqrt{\ 1-\sin\ ^2\theta\ }$$

Therefore cos$$\theta\ =\sqrt{\ 1-\frac{16}{25}}=\sqrt{\ \frac{9}{25}}=\frac{3}{5}$$

$$\tan\ \theta\ =\frac{\sin\theta}{\cos\theta\ \ }\ $$

Therefore $$\tan\theta\ =\frac{4}{3},\cot\theta\ =\frac{3}{4},\sec\theta\ =\frac{5}{3}$$
Now, putting the values in  $$\frac{4\tan\theta-5\cos\theta\ \ }{\sec\theta\ +4\cot\theta\ }$$
Therefore we get $$\frac{\left(\frac{16}{3}-3\right)}{\frac{5}{3}+3}=\ \frac{7}{14}=\frac{1}{2}$$