## Trigonometry (11th Edition) Clone

$$\sin\theta=\frac{2\sqrt6}{5}$$ $$\sec\theta=5$$ $$\csc\theta=\frac{5\sqrt6}{12}$$ $$\tan\theta=2\sqrt6$$ $$\cot\theta=\frac{\sqrt6}{12}$$
$$\cos\theta=\frac{1}{5}\hspace{1.5cm}\theta\hspace{0.1cm}\text{in quadrant I}$$ 1) Reciprocal Identities: $$\sec\theta=\frac{1}{\cos\theta}$$ $$\sec\theta=\frac{1}{\frac{1}{5}}=5$$ 2) Pythagorean Identities: $$\tan^2\theta+1=\sec^2\theta$$ $$\tan^2\theta=\sec^2\theta-1$$ $$\tan^2\theta=5^2-1=25-1=24$$ $$\tan\theta=\pm\sqrt{24}=\pm2\sqrt6$$ 3) Quotient Identities: $$\tan\theta=\frac{\sin\theta}{\cos\theta}$$ We know that $\theta$ is in quadrant I, where both $\sin\theta\gt0$ and $\cos\theta\gt0$. Therefore, with the above identity, we can conclude that $\tan\theta\gt0$. $$\tan\theta=2\sqrt6$$ $$\frac{\sin\theta}{\cos\theta}=2\sqrt6$$ $$\sin\theta=(2\sqrt6)\times\cos\theta=(2\sqrt6)\times\frac{1}{5}=\frac{2\sqrt6}{5}$$ 4) Reciprocal Identities: $$\tan\theta=\frac{1}{\cot\theta}$$ $$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{2\sqrt6}=\frac{\sqrt6}{2\times6}=\frac{\sqrt6}{12}$$ Also, $$\csc\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{2\sqrt 6}{5}}=\frac{5}{2\sqrt 6}=\frac{5\sqrt 6}{12}$$