# Find sec^2 θ - tan^2 θ.

Find $$\displaystyle\frac{1}{{ \sec{\theta}- \tan{\theta}}}+\frac{1}{{ \sec{\theta}+ \tan{\theta}}}$$

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We are given: $$\displaystyle\frac{1}{{ \sec{\theta}- \tan{\theta}}}+\frac{1}{{ \sec{\theta}+ \tan{\theta}}}$$

Rewrite using the LCD which is $$\displaystyle{\left( \sec{\theta}- \tan{\theta}\right)}{\left( \sec{\theta}+ \tan{\theta}\right)}=\frac{1}{{ \sec{\theta}- \tan{\theta}}}{\left(\frac{{ \sec{\theta}+ \tan{\theta}}}{{ \sec{\theta}+ \tan{\theta}}}\right)}+\frac{1}{{ \sec{\theta}+ \tan{\theta}}}{\left(\frac{{ \sec{\theta}- \tan{\theta}}}{{ \sec{\theta}- \tan{\theta}}}\right)}=\frac{{{\left( \sec{\theta}+ \tan{\theta}\right)}+{\left( \sec{\theta}- \tan{\theta}\right)}}}{{{{\sec}^{2}\theta}-{{\tan}^{2}\theta}}}$$

Recall that $$\displaystyle{{\sec}^{2}\theta}={1}+{{\tan}^{2}\theta}$$ so $${{\sec}^{2}\theta}-{{\tan}^{2}\theta}={1}={2}\frac \sec{\theta}{{1}}={2} \sec{\theta}$$