Question

HELP!!! ASAP!!!! TRIG!!!

Answer 1

Answer:

Identity is true

Step-by-step explanation:

$\frac{cos\theta+1}{tan^2\theta}=\frac{cos\theta}{sec\theta-1}$

$(cos\theta+1)(sec\theta-1)=(tan^2\theta)(cos\theta)$

$(cos\theta)(sec\theta)+(cos\theta)(-1)+(1)(sec\theta)+(1)(-1)=(\frac{sin^2\theta}{cos^2\theta})(cos\theta)$

$(cos\theta)(\frac{1}{cos\theta})-cos\theta+sec\theta-1=\frac{sin^2\theta}{cos\theta}$

$1-cos\theta+sec\theta-1=\frac{sin^2\theta}{cos\theta}$

$-cos\theta+sec\theta=\frac{sin^2\theta}{cos\theta}$

$sec\theta-cos\theta=\frac{sin^2\theta}{cos\theta}$

$\frac{1}{cos\theta}-cos\theta=\frac{sin^2\theta}{cos\theta}$

$\frac{1}{cos\theta}-\frac{cos^2\theta}{cos\theta}=\frac{sin^2\theta}{cos\theta}$

$\frac{1-cos^2\theta}{cos\theta}=\frac{sin^2\theta}{cos\theta}$

$\frac{sin^2\theta}{cos\theta}=\frac{sin^2\theta}{cos\theta}$

Therefore, the identity is true.

Helpful tips:

Pythagorean Identity: $sin^2\theta+cos^2\theta=1\\cos^2\theta=1-sin^2\theta\\sin^2\theta=1-cos^2\theta$

Quotient Identities: $tan\theta=\frac{sin\theta}{cos\theta},sec\theta=\frac{1}{cos\theta}$