Question

Prove the following

1.$(1-{\sin }^{2}A)se{c}^{2}A=1$

2.${\sec }^4\theta -{\sec }^2\theta ={\tan }^4\theta +{\tan }^2$

3.$(\sec \theta -\tan \theta {)}^2=\frac{1-\sin \theta }{1+\sin \theta }$

4.$\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +}=\frac{1+\sin \theta }{\cos \theta }$

Answer 1

1. $(1-{\sin }^{2}A){\sec }^{2}A={\cos }^{2}A\cdot \frac{1}{{\cos }^{2}A}$

$=1$

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

$=(1+{\tan }^2\theta )\left({\tan }^2\theta \right)$

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

3. $(\sec \theta -\tan \theta {)}^2=\frac{(1-\sin \theta {)}^2}{{\cos }^2\theta }$

$=\frac{(1-\sin \theta {)}^2}{1-{\sin }^2\theta }$

$=\frac{1-\sin \theta }{1+\sin \theta }$

4. $\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}$

$=\frac{\tan \theta +\sec \theta -1}{\left(\frac{-1}{\sec \theta +\tan \theta }\right)+1}$

$=\frac{(\sec \theta +\tan \theta )(\tan \theta +\sec \theta -1)}{(-1+\sec \theta +\tan \theta )}$

$=\sec \theta +\tan \theta$

$=\frac{1+\sin \theta }{\cos \theta }$