## Trigonometry (11th Edition) Clone

$$\tan\theta+\cot\theta=\sec\theta\csc\theta$$ The proof is below.
$$\tan\theta+\cot\theta=\sec\theta\csc\theta$$ We take a look at the left side first. $$X=\tan\theta+\cot\theta$$ - Quotient identities: $$\tan\theta=\frac{\sin\theta}{\cos\theta}\hspace{2cm}\cot\theta=\frac{\cos\theta}{\sin\theta}$$ Then, $$X=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}$$ $$X=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$$ - Pythagorean identities: $\sin^2\theta+\cos^2\theta=1$ $$X=\frac{1}{\sin\theta\cos\theta}$$ $$X=\frac{1}{\sin\theta}\times\frac{1}{\cos\theta}$$ - Finally, Reciprocal identities: $$\csc\theta=\frac{1}{\sin\theta}\hspace{2cm}\sec\theta=\frac{1}{\cos\theta}$$ Thus, $$X=\csc\theta\sec\theta$$ So 2 sides are equal. The equation is an identity as a result.