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3 years ago

13

Help me asap please!!!!!

2 answers:

faust18 [17]3 years ago

7 0

Tuberculosis I believe is right

belka [17]3 years ago

3 0

Tuberculosis is the answer to that question

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9.<br>(1 + tan²theta) cot theta<br>cosec theta<br>tan theta

wolverine [178]

Answer:

Proved

Explanation:

Given

$\frac{(1+tan^2\theta). cot\theta}{cosec^2 \theta}= tan \theta$

Required

Prove

In trigonometry:

$1 + tan^2\theta = sec^2\theta$

So, we have:

$\frac{sec^2\theta * cot\theta}{cosec^2 \theta}= tan \theta$

Express each identity as an inverse:

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

Rewrite as:

$(\frac{1}{cos^2\theta} * \frac{1}{tan\theta})/ \frac{1}{sin^2 \theta}= tan \theta$

Express tan as sin/cos:

$(\frac{1}{cos^2\theta} * \frac{1}{sin\theta/cos\theta})/ \frac{1}{sin^2 \theta}= tan \theta$

$(\frac{1}{cos^2\theta} * \frac{1}{sin\theta/cos\theta}) * sin^2 \theta= tan \theta$

$(\frac{1}{cos^2\theta} * \frac{cos\theta}{sin\theta}) * sin^2 \theta= tan \theta$

$(\frac{1}{cos\theta} * \frac{1}{sin\theta}) * sin^2 \theta= tan \theta$

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

$\frac{sin \theta}{cos\theta}= tan \theta$

$tan \theta= tan \theta$

Proved

3 0

3 years ago