Question

If ${\cos }^4\theta +{\cos }^2\theta =1$ then show that
i) ${\sec }^4\theta -{\sec }^2\theta =1$
ii) ${\cot }^4\theta -{\cot }^2\theta =1$

Answer 1

Given: ${\cos }^4\theta +{\cos }^2\theta =1$

${\cos }^4\theta =1-{\cos }^2\theta$

${\cos }^4\theta ={\sin }^2\theta \left(1\right)$

Now,

$\left(1\right){\sec }^4\theta -{\sec }^2\theta$

$=\frac{1}{{\cos }^4\theta }-\frac{1}{{\cos }^2\theta }$

$=\frac{1-{\cos }^2\theta }{{\cos }^4\theta }$

$=\frac{{\sin }^2\theta }{{\cos }^4\theta }$

$=\frac{{\sin }^2\theta }{{\sin }^2\theta }\left(from\left(1\right)\right)$

$=1$

$\left(2\right){\cot }^4\theta -{\cot }^2\theta$

$=\frac{{\cos }^4\theta }{{\sin }^4\theta }-\frac{{\cos }^2\theta }{{\sin }^2\theta }$

$=\frac{{\cos }^4\theta -{\sin }^2\theta {\cos }^2\theta }{{\sin }^4\theta }$

$=\frac{{\sin }^2\theta -{\sin }^2\theta {\cos }^2\theta }{{\sin }^4\theta }\left[from\left(1\right)\right]$

$=\frac{{\sin }^2\theta \left(1-{\cos }^2\theta \right)}{{\sin }^4\theta }$

$=\frac{{\sin }^2\theta \times {\sin }^2\theta }{{\sin }^4\theta }$

$=\frac{{\sin }^4\theta }{{\sin }^4\theta }$

$=1$