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djyliett [7]
4 years ago
9

Prove: cos⁡(x)/(1+sin⁡(x) )+(1+sin⁡(x))/cos⁡(x) =2sec⁡(x)

Mathematics
2 answers:
weeeeeb [17]4 years ago
6 0
\dfrac{\cos x}{1+\sin x}+\dfrac{1+\sin x}{\cos x}=2\sec x\\\\L_s=\dfrac{\cos x\cos x}{\cos x(1+\sin x)}+\dfrac{(1+\sin x)(1+\sin x)}{\cos x(1+\sin x)}\\\\=\dfrac{\cos^2x+1+\sin x+\sin x+\sin^2x}{\cos x(1+\sin x)}=\dfrac{(\cos^2x+\sin^2x)+1+2\sin x}{\cos x(1+\sin x)}\\\\=\dfrac{1+1+2\sin x}{\cos x(1+\sin x)}=\dfrac{2+2\sin x}{\cos x(1+\sin x)}=\dfrac{2(1+\sin x)}{\cos x(1+\sin x)}\\\\=\dfrac{2}{\cos x}=2\cdot\dfrac{1}{\cos x}=2\sec x=R_s

\text{Used:}\\\\\sin^2\alpha+\cos^2\alpha=1\\\\\sec\alpha=\dfrac{1}{\cos\alpha}
saw5 [17]4 years ago
3 0
<span>cos⁡(x)/(1+sin⁡(x) )+(1+sin⁡(x))/cos⁡(x) =2sec⁡(x)

Work on the left hand side.
</span>[Common denominator is (1+sin(x))*cos(x)]
cos⁡(x)/(1+sin⁡(x) )+(1+sin⁡(x))/cos⁡(x)    <span>
= (cos(x)^2+(1+sin(x))^2)/(</span>(1+sin(x))*cos(x))
=(cos(x)^2+1+sin(x)^2+2sin(x))/((1+sin(x))*cos(x))
=(cos(x)^2+sin(x)^2+1+2sin(x))/((1+sin(x))*cos(x))
=(1+1+2sin(x))/((1+sin(x))*cos(x))
=(2+2sin(x))/((1+sin(x))*cos(x))
=2(1+sin(x))/((1+sin(x))*cos(x))
=2/cos(x)
=2 sec(x)                [QED]

<span>

</span>
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Step-by-step explanation:

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8 0
3 years ago
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what's the error? MAX said that 36,594,145 is less than 5,980,251 because 3 is less than 5. Describe max error and give the corr
forsale [732]
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3 years ago
Find a formula for the sum of the first n even positive integers.
melisa1 [442]
So even postive integers are by defention in form 2k where k is a natural number so
let the sum of even integers to n=S
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divide bith sides of equation 1 by 2
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divide both sides of equation 2 by 2
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by adding both we will get
___________________________
S=(k+1)(k)
so the sum will be equal to
S=
{k}^{2}  + k
so let us test the equation
for the first 3 even number there sums will be
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Answer:

Angle parking is more common than perpendicular parking.

Angle parking spots have half the blind spot as compared to perpendicular parking spaces

Step-by-step explanation:

Considering the available options, the true statement about angle parking is that" Angle parking is more common than perpendicular parking." Angle parking is mostly constructed and used for public parking. It is mostly used where the parking lots are quite busy such as motels or public garages.

Therefore, in this case, the answer is that "Angle parking is more common than perpendicular parking."

Also, "Angle parking spots have half the blind spot as compared to perpendicular parking spaces."

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