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Svetach [21]
3 years ago
7

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

Mathematics
1 answer:
RSB [31]3 years ago
6 0

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.

<u>Helpful tips:</u>

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}

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