Given that tan θ = 5/4, we can use the identity \(\tan^2\theta + 1 = \sec^2\theta\) to find sec θ:

\(\sec^2\theta = \tan^2\theta + 1 = \left(\frac{5}{4}\right)^2 + 1 = \frac{25}{16} + 1 = \frac{41}{16}\)

Taking the square root of both sides, we get:

\(\sec\theta = \pm\sqrt{\frac{41}{16}}\)

Since sec θ is the reciprocal of cos θ, we have:

\(\cos\theta = \pm\frac{4}{\sqrt{41}}\)

We can use the identity \(\sin^2\theta + \cos^2\theta = 1\) to find sin θ:

\(\sin^2\theta = 1 - \cos^2\theta = 1 - \left(\pm\frac{4}{\sqrt{41}}\right)^2 = 1 - \frac{16}{41} = \frac{25}{41}\)

Taking the square root of both sides, we get:

\(\sin\theta = \pm\frac{5}{\sqrt{41}}\)

Substituting these values of sin θ and cos θ into the given expression, we get:

\(\sin^2\theta - \cos^2\theta = \left(\pm\frac{5}{\sqrt{41}}\right)^2 - \left(\pm\frac{4}{\sqrt{41}}\right)^2 = \frac{25}{41} - \frac{16}{41} = \frac{9}{41}\)

So, if tan θ=5/4, then the value of \(\sin^2θ-\cos^2θ\) is **9/41**.