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Airida [17]
3 years ago
12

Probe that:

29%20%7D%20%20%5Calpha%20%20%3D%201" id="TexFormula1" title=" \sec \alpha \sqrt{1 - \sin( {}^{2} ) } \alpha = 1" alt=" \sec \alpha \sqrt{1 - \sin( {}^{2} ) } \alpha = 1" align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Nataly [62]3 years ago
7 0

Step-by-step explanation:

<h3>\sec \alpha  \sqrt{1 -  \sin ^{2}   \alpha }  = 1</h3>

Prove the LHS

Using trigonometric identities

That's

<h3>\cos ^{2}  \alpha  = 1 -  \sin^{2}  \alpha</h3>

<u>Rewrite the expression</u>

We have

<h3>\sec \alpha  \sqrt{ \cos^{2} \alpha  }</h3>

<h3>\sqrt{ { \cos }^{2}  \alpha }  =  \cos \alpha</h3>

So we have

<h3>\sec  \alpha  \times  \cos \alpha</h3>

Using trigonometric identities

<h3>\sec \alpha  =  \frac{1}{ \cos \alpha }</h3>

<u>Rewrite the expression</u>

That's

<h3>\frac{1}{\cos \alpha }  \times  \cos \alpha</h3>

Reduce the expression with cos a

We have the final answer as

<h2>1</h2>

As proven

Hope this helps you

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Answer:

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~

Step-by-step explanation:

Let the side of square be 's'. We know that Perimeter of square is <u>4 ×</u> <u>Side</u>.<em> </em>By substituting values in the formula we can find the required side.

~

:\implies \sf  4 × s=Perimeter \:  of \:  square

~

:\implies \sf  \:  \:  \: 4 × s = 12

~

:\implies \sf \:  \:  \:  s =  \dfrac{12}{4}

~

:\implies \bf \:  \:  \:  Side = 3 yd.

~

\therefore \: Length of side of square is <u>3</u><u> </u><u>yd</u>.

7 0
1 year ago
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the answer is
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8 0
2 years ago
Read 2 more answers
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