Question

$\sec \left(\frac{3\pi }{2}-\theta \right)\sec \left(\theta -\frac{5\pi }{2}\right)+\tan \left(\frac{5\pi }{2}+\theta \right)\tan \left(\theta -\frac{3\pi }{2}\right)$
is equal to

• A.
  0
• B.
  -1
• C.
  1
• D.
  2

Answer 1

The correct option is B

-1
We first try to simplify the expression
Upon simplifying we get,
$\sec \left(\frac{3\pi }{2}-\theta \right)\sec \left(\theta -\frac{5\pi }{2}\right)+\tan \left(\frac{5\pi }{2}+\theta \right)\tan \left(\theta -\frac{3\pi }{2}\right)=\sec \left(\frac{3\pi }{2}-\theta \right)\sec \left(\frac{5\pi }{2}-\theta \right)-\tan \left(\frac{5\pi }{2}+\theta \right)\tan \left(\frac{3\pi }{2}-\theta \right)$
Also,
$\frac{5\pi }{2}=2\pi +\frac{\pi }{2}$

Thus, we can write
$\sec \left(\frac{5\pi }{2}-\theta \right)=\sec \left(2\pi +\frac{\pi }{2}-\theta \right)\Rightarrow \sec \left(\frac{5\pi }{2}-\theta \right)=\sec \left(\frac{\pi }{2}-\theta \right)\mathrm{and}\mathrm{similarly}\tan \left(\frac{5\pi }{2}-\theta \right)=\tan \left(\frac{\pi }{2}-\theta \right)$
Thus, our expression becomes
$\sec \left(\frac{3\pi }{2}-\theta \right)\sec \left(\frac{\pi }{2}-\theta \right)-\tan \left(\frac{3\pi }{2}-\theta \right)\tan \left(\frac{\pi }{2}-\theta \right)$
Now, using the identities for complimentary angles
$\sec \left(\frac{3\pi }{2}-\theta \right)=-\mathrm{cosec}\theta sec\left(\frac{\pi }{2}-\theta \right)=\mathrm{cosec}\theta and\tan \left(\frac{3\pi }{2}-\theta \right)=-\mathrm{co}t\theta t\mathrm{an}\left(\frac{\pi }{2}-\theta \right)=\mathrm{co}t\theta$
we get our expression as:
$-\cos e{c}^2\theta +co{t}^2\theta =-(\cos e{c}^2\theta -co{t}^2\theta )=-1$
Hence, Option b is correct.