Lease please help me I will give brainliest to the first right answer!!!!

Mathematics

2 answers:

Alexus [3.1K]3 years ago

8 0

Answer:

a.option is the correct answer right I think so

Olin [163]3 years ago

4 0

Answer:

A seems to be the right answer

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What is an equation of the line that passes through the points (-5, 8) and (5,0)?

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3 years ago

Simplify cotø(tanø+cotø)

Sladkaya [172]

Answer:

$\large\boxed{\cot\theta(\tan\theta+\cot\theta)=1+\cot^2\theta=\dfrac{1}{\sin^2\theta}=\csc^2\theta}$

Step-by-step explanation:

$\text{Use}\\\\\text{distributive property:}\ a(b+c)=ab+ac\\\cot\alpha\tan\alpha=1.\\\\======================\\\\\cot\theta(\tan\theta+\cot\theta)=(\cot\theta)(\tan\theta)+(\cot\theta)(\cot\theta)\\\\=1+\cot^2\theta\\\\\text{If you want next transformation, then use:}\\\\\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\\\\\sin^2\alpha+\cos^2\alpha=1\\\\=======================$

$=1+\left(\dfrac{\cos\theta}{\sin\theta}\right)^2=1+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta}{\sin^2\theta}+\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}\\\\=\dfrac{1}{\sin^2\theta}\\\\\text{If you want next transformation, then use:}\\\\\csc\alpha=\dfrac{1}{\sin\alpha}\\\\=\left(\dfrac{1}{\sin\theta}\right)^2=(\csc\theta)^2=\csc^2\theta$