## Trigonometry (11th Edition) Clone

$$\sin\theta+\cos\theta=\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$$ The expression has been proved to be an identity by simplifying the right side.
$$\sin\theta+\cos\theta=\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$$ The right side looks more complex, so we would deal with it first. $$A=\frac{\sin\theta}{1-\cot\theta}+\frac{\cos\theta}{1-\tan\theta}$$ We would transform $\cot\theta$ and $\tan\theta$: $$\cot\theta=\frac{\cos\theta}{\sin\theta}\hspace{2cm}\tan\theta=\frac{\sin\theta}{\cos\theta}$$ Therefore, $A$ would be $$A=\frac{\sin\theta}{1-\frac{\cos\theta}{\sin\theta}}+\frac{\cos\theta}{1-\frac{\sin\theta}{\cos\theta}}$$ $$A=\frac{\sin\theta}{\frac{\sin\theta-\cos\theta}{\sin\theta}}+\frac{\cos\theta}{\frac{\cos\theta-\sin\theta}{\cos\theta}}$$ $$A=\frac{\sin^2\theta}{\sin\theta-\cos\theta}+\frac{\cos^2\theta}{\cos\theta-\sin\theta}$$ $$A=\frac{\sin^2\theta}{\sin\theta-\cos\theta}-\frac{\cos^2\theta}{\sin\theta-\cos\theta}$$ (we change the sign of the numerator since the sign of the denominator has changed) $$A=\frac{\sin^2\theta-\cos^2\theta}{\sin\theta-\cos\theta}$$ $$A=\frac{(\sin\theta-\cos\theta)(\sin\theta+\cos\theta)}{\sin\theta-\cos\theta}$$ (for $a^2-b^2=(a-b)(a+b)$) $$A=\sin\theta+\cos\theta$$ The left side is therefore equal with the right side. The expression has been proved to be an identity.