We should proceed with the fact that sum of angles on one side of a straight line is $180^\circ$
So from the given figure,
$\theta+\phi+90^\circ=180^\circ$
So, $\theta=90^\circ-\phi\ \dots(1)$
Now from the triangle $\triangle\text{ABC},$
$\sin\theta=\frac{4}{5}$
Now we will use equation (1) in the above,
$\sin(90^\circ-\phi)=\frac{4}{5}$
Therefore, $\cos\phi=\frac{4}{5}$
So the answer is $(d)$
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If $\theta$ is an acute angle such that $\sec^2\theta=3,$ then the value of $\frac{\tan^2\theta-\text{cosec}^2\theta}{\tan^2\theta+\text{cosec}^2\theta}$ is: