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3 years ago

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Verifying Trig Functions

1 answer:

anzhelika [568]3 years ago

5 0

Answer:

The answer to your question is below

Step-by-step explanation:

csc α - cot α = $\frac{sin \alpha }{1 + cos \alpha }$

Work with the left part

$\frac{1}{sin\alpha } - \frac{cos \alpha }{sin \alpha}$

Simplify

$\frac{1 - cos\alpha }{sin\alpha }$

Multiply by the reciprocal

$\frac{1 - cos\alpha }{sin \alpha } x \frac{1 + cos\alpha }{1 + cos\alpha }$

Simplify

$\frac{1 - cos^{2}\alpha}{sin\alpha (1 + cos\alpha)}$

Remember that

sin²α = 1 -cos²α

Substitute the previous equation

$\frac{sin^{2}\alpha }{sin\alpha(1 + cos\alpha ) }$

Simplify

$\frac{sin\alpha}{1 + cos \alpha}$

The identity i proved.

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the sum of two numbers is 100. the sum of 8 and the larger number is equal to 5 times the smaller number. What is the smaller nu

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Answer:

x=82

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Step-by-step explanation:

x+y=100

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saul85 [17]

$\large\underline{\sf{Solution-}}$

<u>Let us assume that:</u>

$\sf \longmapsto x = \sqrt{6 + \sqrt{6 + \sqrt{6 + ... \infty } } }$

We can also write it as:

$\sf \longmapsto x = \sqrt{6 + x }$

Squaring both sides, we get:

$\sf \longmapsto {x}^{2} =6 + x$

$\sf \longmapsto {x}^{2} - x - 6 =0$

By splitting the middle term:

$\sf \longmapsto {x}^{2} - 3x + 2x - 6 =0$

$\sf \longmapsto x(x - 3) + 2(x - 3 ) =0$

$\sf \longmapsto (x+ 2)(x - 3 ) =0$

<u>Therefore:</u>

$\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}$

$\sf \longmapsto x = - 2,3$

<u>But x cannot be negative. </u>

$\sf \longmapsto x = 3$

Therefore, the value of the expression is 3.

$\large\underline{\sf{Verification-}}$

Given:

$\sf\longmapsto x=3$

We can also write it as:

$\sf\longmapsto x = \sqrt{9}$

$\sf\longmapsto x = \sqrt{6+3}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{9}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}$

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}$

This pattern will continue.

$\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}$

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