Trigonometry (11th Edition) Clone

The equation $$\cot\theta\tan(\theta+\pi)-\sin(\pi-\theta)\cos\Big(\frac{\pi}{2}-\theta\Big)=\cos^2\theta$$ is verified to be an identity, as proved in the Work Step by Step.
$$\cot\theta\tan(\theta+\pi)-\sin(\pi-\theta)\cos\Big(\frac{\pi}{2}-\theta\Big)=\cos^2\theta$$ The left side would be examined first. $$X=\cot\theta\tan(\theta+\pi)-\sin(\pi-\theta)\cos\Big(\frac{\pi}{2}-\theta\Big)$$ - This exercise asks for the application of the identities of tangent sum, sine difference and cosine difference, which state $$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$ $$\sin(A-B)=\sin A\cos B-\cos A\sin B$$ $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ Expand $\tan(\theta+\pi)$, $\sin(\pi-\theta)$ and $\cos\Big(\frac{\pi}{2}-\theta\Big)$ in terms of the identities, we get $$X=\cot\theta\times\frac{\tan\theta+\tan\pi}{1-\tan\theta\tan\pi}-(\sin\pi\cos\theta-\cos\pi\sin\theta)\times\Big(\cos\frac{\pi}{2}\cos\theta+\sin\frac{\pi}{2}\sin\theta\Big)$$ Recall that $\tan\pi=0$, $\sin\pi=0$, $\cos\pi=-1$, $\cos\frac{\pi}{2}=0$ and $\sin\frac{\pi}{2}=1$. $$X=\cot\theta\times\frac{\tan\theta+0}{1-\tan\theta\times0}-[0\times\cos\theta-(-1)\sin\theta]\times(0\times\cos\theta+1\times\sin\theta)$$ $$X=\cot\theta\times\frac{\tan\theta}{1}-[\sin\theta]\times(\sin\theta)$$ $$X=\cot\theta\tan\theta-\sin^2\theta$$ - From Reciprocal Identities, we get that $\cot\theta=\frac{1}{\tan\theta}$. Thus, $\cot\theta\tan\theta=1$. $$X=1-\sin^2\theta$$ - From Pythagorean Identities, we also get that $1-\sin^2\theta=\cos^2\theta$. Thus, $$X=\cos^2\theta$$ Therefore, we can conclude now that the equation $$\cot\theta\tan(\theta+\pi)-\sin(\pi-\theta)\cos\Big(\frac{\pi}{2}-\theta\Big)=\cos^2\theta$$ is verified to be an identity.