Question

If $co{s}^2\Theta +co{s}^4\Theta =1$ show that $ta{n}^2\Theta +ta{n}^4\Theta =1$

Answer 1

We have,

$co{s}^2\theta +co{s}^4\theta =1$

$=co{s}^4\theta =1-co{s}^2\theta$

$=co{s}^4\theta =si{n}^2\theta$

Dividing both sides by $co{s}^2\theta$ we get,

$co{s}^2\theta =\frac{si{n}^2\theta }{co{s}^2\theta }$

$co{s}^2\theta =ta{n}^2\theta$

And, $ta{n}^2\theta +ta{n}^4\theta =\left(ta{n}^2\theta \right)(1+ta{n}^2\theta )$

$=ta{n}^2\theta se{c}^2\theta$

$=\frac{ta{n}^2\theta }{co{s}^2\theta }$

$=\frac{co{s}^2\theta }{co{s}^2\theta }$

$=1$