## Trigonometry (11th Edition) Clone

$$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$ The step of verifying trigonometric identity is as below.
$$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$ We start from the left side first. $$X=1-\tan^2\frac{\theta}{2}$$ - Half-angle identity for tangent: $$\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$$ Thus, $$\tan^2\frac{\theta}{2}=\frac{\sin^2\theta}{(1+\cos\theta)^2}$$ Therefore, $$X=1-\frac{\sin^2\theta}{(1+\cos\theta)^2}$$ $$X=\frac{(1+\cos\theta)^2-\sin^2\theta}{(1+\cos\theta)^2}$$ $$X=\frac{1+2\cos\theta+\cos^2\theta-\sin^2\theta}{(1+\cos\theta)^2}$$ (we know that $(A+B)^2=A^2+2AB+B^2$) - Now we need to rewrite $\sin^2\theta$ into $1-\cos^2\theta$, as represented in Pythagorean identity. $$X=\frac{1+2\cos\theta+\cos^2\theta-(1-\cos^2\theta)}{(1+\cos\theta)^2}$$ $$X=\frac{1+2\cos\theta+\cos^2\theta-1+\cos^2\theta}{(1+\cos\theta)^2}$$ $$X=\frac{2\cos\theta+2\cos^2\theta}{(1+\cos\theta)^2}$$ $$X=\frac{2\cos\theta(1+\cos\theta)}{(1+\cos\theta)^2}$$ $$X=\frac{2\cos\theta}{1+\cos\theta}$$ Finally, $$1-\tan^2\frac{\theta}{2}=\frac{2\cos\theta}{1+\cos\theta}$$ The equation has been verified to be an identity.