## Trigonometry (11th Edition) Clone

$$\sec(\pi-x)=-\sec x$$ The equation has been verified to be an identity as below.
$$\sec(\pi-x)=-\sec x$$ The left side should be rewritten first as it is more complex. $$X=\sec(\pi-x)$$ - From reciprocal identity: $$\sec\theta=\frac{1}{\cos\theta}$$ Therefore, $$X=\frac{1}{\cos(\pi-x)}$$ - From difference identity for cosine: $\cos(A-B)=\cos A\cos B+\sin A\sin B$ Apply the identity for $\cos(\pi-x)$: $$X=\frac{1}{\cos \pi\cos x+\sin \pi\sin x}$$ $$X=\frac{1}{-1\times\cos x+0\times\sin x}$$ $$X=\frac{1}{-\cos x}$$ $$X=-\frac{1}{\cos x}$$ Apply back the reciprocal identity mentioned in the beginning to rewrite $\frac{1}{\cos x}$ into $\sec x$. $$X=-\sec x$$ So, it has been proved that $$\sec(\pi-x)=-\sec x$$ and the equation is an identity.