Question

verify that the equation is an identity
\\( \frac { \cos \theta + 1 } { \tan ^ { 2 } \theta } = \frac { \cos \theta } { \sec \theta - 1 } \\)
to verify the identity, start with the more complicated side and transform it to look like the other side. choose the correct transformations and transfor
\\( \frac { \cos \theta + 1 } { \tan ^ { 2 } \theta } \\)
\\( = \frac { \cos \theta + 1 } { \sec ^ { 2 } \theta - 1 } \\)
\\( \tan ( - \theta ) = - \tan \theta \\)
\\( \tan \theta = \frac { 1 } { \cot \theta } \\)
\\( \tan ^ { 2 } \theta + 1 = \sec ^ { 2 } \theta \\)
\\( \tan \theta = \frac { \sin \theta } { \cos \theta } \\)

Explanation:

Step1: Use the Pythagorean identity

We know the Pythagorean identity \( \tan^{2}\theta+1 = \sec^{2}\theta\), which can be rewritten as \( \tan^{2}\theta=\sec^{2}\theta - 1\). So, \(\frac{\cos\theta + 1}{\tan^{2}\theta}=\frac{\cos\theta + 1}{\sec^{2}\theta - 1}\)

Step2: Factor the denominator

Since \(a^{2}-b^{2}=(a + b)(a - b)\), and \(\sec^{2}\theta-1=(\sec\theta + 1)(\sec\theta - 1)\). Then \(\frac{\cos\theta + 1}{\sec^{2}\theta - 1}=\frac{\cos\theta + 1}{(\sec\theta + 1)(\sec\theta - 1)}\)

Step3: Use the reciprocal identity \(\sec\theta=\frac{1}{\cos\theta}\)

Multiply the numerator and denominator by \(\cos\theta\). The numerator becomes \(\cos\theta(\cos\theta + 1)=\cos^{2}\theta+\cos\theta\), and the denominator becomes \((\frac{1}{\cos\theta}+ 1)(\sec\theta - 1)\cos\theta=(1+\cos\theta)(\sec\theta - 1)\)

Step4: Simplify the fraction

Cancel out the common factor \((\cos\theta + 1)\) (assuming \(\cos\theta
eq - 1\)). We get \(\frac{\cos\theta}{\sec\theta - 1}\)

Answer:

The given equation \(\frac{\cos\theta + 1}{\tan^{2}\theta}=\frac{\cos\theta}{\sec\theta - 1}\) is an identity.