verify that the equation is an identity.\n\\( \\frac { 2 } { 1 + \\cos x } - \\tan ^ { 2 } \\frac { x } { 2…

verify that the equation is an identity.\n\\( \\frac { 2 } { 1 + \\cos x } - \\tan ^ { 2 } \\frac { x } { 2 } = 1 \\)\nto verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformations and transform the expression at each step.\n\\( \\frac { 2 } { 1 + \\cos x } - \\tan ^ { 2 } \\frac { x } { 2 } = \\frac { 2 } { 1 + \\cos x } - \\square \\)\napply a pythagorean identity.\ndivide each term by tangent.\napply a reciprocal identity.\napply a half - angle identity.
Answer
Explanation:
Step1: Apply a Pythagorean identity
Recall the Pythagorean identity (1+\tan^{2}\alpha=\sec^{2}\alpha). So, (\frac{2}{1 + \cos x}-\tan^{2}\frac{x}{2}=\frac{2}{1 + \cos x}-\left(\sec^{2}\frac{x}{2}-1\right))
Step2: Apply a reciprocal identity
Since (\sec\alpha=\frac{1}{\cos\alpha}), then (\sec^{2}\frac{x}{2}=\frac{1}{\cos^{2}\frac{x}{2}}). Also, use the half - angle formula (\cos\frac{x}{2}=\sqrt{\frac{1 + \cos x}{2}}), so (\cos^{2}\frac{x}{2}=\frac{1+\cos x}{2}) and (\sec^{2}\frac{x}{2}=\frac{2}{1 + \cos x})
Substitute (\sec^{2}\frac{x}{2}=\frac{2}{1 + \cos x}) into (\frac{2}{1 + \cos x}-\left(\sec^{2}\frac{x}{2}-1\right)):
[ \begin{align*} \frac{2}{1 + \cos x}-\left(\frac{2}{1 + \cos x}-1\right)&=\frac{2}{1 + \cos x}-\frac{2}{1 + \cos x}+1\ &=1 \end{align*} ]
Answer:
The given equation (\frac{2}{1+\cos x}-\tan^{2}\frac{x}{2} = 1) is an identity.