## Trigonometry (11th Edition) Clone

$$\sec(\pi-x)=-\sec x$$ We attempt the left side first to reach the conclusion that the equation is an identity.
$$\sec(\pi-x)=-\sec x$$ We attempt from the left side first. $$A=\sec(\pi-x)$$ It is known that $$\sec\theta=\frac{1}{\cos\theta}$$ which means $$A=\frac{1}{\cos(\pi-x)}$$ Now we expand $\cos(\pi -x)$ according to the cosine difference identity, which states $$\cos(A-B)=\cos A\cos B+\sin A\sin B$$ $$A=\frac{1}{\cos\pi\cos x+\sin\pi\sin x}$$ $$A=\frac{1}{-1\times\cos x+0\times\sin x}$$ $$A=\frac{1}{-\cos x}$$ $$A=-\frac{1}{\cos x}$$ $$A=-\sec x$$ (as $\sec\theta=\frac{1}{\cos\theta}$) The equation is true, so it is an identity.