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Answer:

Identity is true

Step-by-step explanation:

[tex]\frac{cos\theta+1}{tan^2\theta}=\frac{cos\theta}{sec\theta-1}[/tex]

[tex](cos\theta+1)(sec\theta-1)=(tan^2\theta)(cos\theta)[/tex]

[tex](cos\theta)(sec\theta)+(cos\theta)(-1)+(1)(sec\theta)+(1)(-1)=(\frac{sin^2\theta}{cos^2\theta})(cos\theta)[/tex]

[tex](cos\theta)(\frac{1}{cos\theta})-cos\theta+sec\theta-1=\frac{sin^2\theta}{cos\theta}[/tex]

[tex]1-cos\theta+sec\theta-1=\frac{sin^2\theta}{cos\theta}[/tex]

[tex]-cos\theta+sec\theta=\frac{sin^2\theta}{cos\theta}[/tex]

[tex]sec\theta-cos\theta=\frac{sin^2\theta}{cos\theta}[/tex]

[tex]\frac{1}{cos\theta}-cos\theta=\frac{sin^2\theta}{cos\theta}[/tex]

[tex]\frac{1}{cos\theta}-\frac{cos^2\theta}{cos\theta}=\frac{sin^2\theta}{cos\theta}[/tex]

[tex]\frac{1-cos^2\theta}{cos\theta}=\frac{sin^2\theta}{cos\theta}[/tex]

[tex]\frac{sin^2\theta}{cos\theta}=\frac{sin^2\theta}{cos\theta}[/tex]

Therefore, the identity is true.

Helpful tips:

Pythagorean Identity: [tex]sin^2\theta+cos^2\theta=1\\cos^2\theta=1-sin^2\theta\\sin^2\theta=1-cos^2\theta[/tex]

Quotient Identities: [tex]tan\theta=\frac{sin\theta}{cos\theta},sec\theta=\frac{1}{cos\theta}[/tex]

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