Question

Show that $\sec (\frac{3\pi }{2}-\theta )\sec (\theta -\frac{5\pi }{2})+\tan (\frac{5\pi }{2}+\theta )\tan (\theta -\frac{3\pi }{2})=-1$.

Answer 1

$\sec (\frac{3\pi }{2}-\theta )\sec (\theta -\frac{5\pi }{2})+\tan (\frac{5\pi }{2}+\theta )\tan (\theta -\frac{3\pi }{2})$

simplifying

$=(-\csc \theta )\left(\sec (2\pi +\frac{\pi }{2}-\theta )\right)+\tan (2\pi +\frac{\pi }{2}+\theta )\times (-\tan (\frac{3\pi }{2}-\theta ))$

$=-\csc \theta \left(\sec (\frac{\pi }{2}-\theta )\right)+\tan (\frac{\pi }{2}+\theta )(-\cot \theta )=-\csc \theta \csc \theta +{\cot }^2\theta$

$=-\frac{1}{{\sin }^2\theta }+\frac{{\cos }^2\theta }{{\sin }^2\theta }=-\frac{1+{\cos }^2\theta }{{\sin }^2\theta }=-\frac{{\sin }^2\theta }{{\sin }^2\theta }=-1$