If tanθ - cotθ = 0°, 0° < θ < 90° then, find the value of sin2θ + sec…

tanθ - cotθ = 0° (only possible when tanθ = cotθ)

So, the θ must be 45°.

So,

sin²θ + sec²θ = (\(\frac{1}{\sqrt {2}}\))²+(\(\sqrt {2}\))²

= \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)