Question

If ${\cos }^4\theta +{\cos }^2\theta =1$ then show that
(i) $se{c}^4\theta -se{c}^2\theta =1$
(ii) ${\cot }^4\theta -co{t}^2\theta =1$
(iii) $ta{n}^4\theta -ta{n}^2\theta =1$

Answer 1

(i) ${\cos }^4\theta +{\cos }^2\theta =1$

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

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

Now,

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

$={\sec }^2\theta {\tan }^2\theta =\frac{{\sin }^2\theta }{{\cos }^4\theta }=1$

(ii) ${\cot }^4\theta -2{\cot }^2\theta =1$

We have to show

${\cot }^4\theta =1+{\cot }^2\theta =\cos e{c}^2\theta$

Now, L.H.S.

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

$=\cos e{c}^2\theta$

R.H.S

(iii) ${\tan }^4\theta -{\tan }^2\theta =1$

We need to show

${\tan }^4\theta =1+{\tan }^2\theta$

$={\sec }^2\theta$

Now, ${\tan }^4\theta =\frac{{\sin }^4\theta }{{\cos }^4\theta }$

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

$={\sin }^2\theta$