If $$tan\theta$$ = 3/4 and $$\theta$$ is acute, then $$cosec\theta$$ is equal to
$$\frac{sin \theta}{cos \theta} = \frac{3}{4}$$
So, $$\frac{sin^2\theta}{cos^2\theta}=\frac{9}{16}$$
Hence, $$sin^2 \theta = \frac{9}{9+16}=\frac{9}{25}$$
So, $$cosec \theta = \frac{5}{3}$$
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