Trigonometry (11th Edition) Clone

$$(1-\cos\theta)(1+\sec\theta)=\sin\theta\tan\theta$$
$$A=(1-\cos\theta)(1+\sec\theta)$$ - Reciprocal Identity: $$\sec\theta=\frac{1}{\cos\theta}$$ Replace into $A$: $$A=(1-\cos\theta)(1+\frac{1}{\cos\theta})$$ $$A=(1-\cos\theta)(\frac{\cos\theta+1}{\cos\theta})$$ $$A=\frac{(1-\cos\theta)(1+\cos\theta)}{\cos\theta}$$ As $(a-b)(a+b)=a^2-b^2$: $$A=\frac{1-\cos^2\theta}{\cos\theta}$$ - Pythagorean Identity: $$\sin^2\theta=1-\cos^2\theta$$ Replace into $A$: $$A=\frac{\sin^2\theta}{\cos\theta}$$ As the exercise requests no quotients appear in the final expression, we continue. $$A=\sin\theta\times\frac{\sin\theta}{\cos\theta}$$ $$A=\sin\theta\tan\theta\hspace{1cm}\text{(Quotient Identity)}$$