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1 year ago

Sin10°-cos10° product and evaluate

Mathematics

2 answers:

Elena-2011 [213]1 year ago

8 0

$\\ \sf\longmapsto sin10°-cos10°$

$\\ \sf\longmapsto sin(90-10°)-cos10°$

sin(90-A)=cosA

$\\ \sf\longmapsto cos10-cos10$

$\\ \sf\longmapsto 0$

Montano1993 [528]1 year ago

6 0

Question should be sin 20 cos 10 + cos 20 sin 10

remember sin (a+b) = sin a cos b + cos a sin b ?

sin 20 cos 10 + cos 20 sin 10

= sin (20+10)

= sin 30

= 1/2

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1. 5x^2 - 3

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2 years ago

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Alchen [17]

Answer:

$y=-2x+12$

Step-by-step explanation:

We can use the point-slope form given by:

$y-y_1=m(x-x_1)$

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So, we will substitute -2 for <em>m</em> and (5, 2) for (x₁, y₁). This gives us:

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Add 2 to both sides. Hence, our equation is:

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2 years ago

Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A

iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

$\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA$

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2 years ago

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2 years ago

Sin10°-cos10° product and evaluate​

Sin10°-cos10° product and evaluate