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N76 [4]
3 years ago
11

Does anyone know how to solve this?

Mathematics
1 answer:
MrRissso [65]3 years ago
5 0

9514 1404 393

Answer:

Step-by-step explanation:

The tangent of the angle can be found using the identity for the tangent of the difference of angles.

  tan(α-β) = (tan(α) -tan(β))/(1 +tan(α)tan(β))

The tangent of the angle of each vector is the ratio of the j coefficient to the i coefficient.

  tan(α) = -2/3

  tan(β) = 4/7

  tan(α-β) = (-2/3 -4/7)/(1 +(-2/3)(4/7))

  = (-14-12)/(21 -8) = -26/13 = -2

Then the magnitude of the angle between the vectors is ...

  arctan(2) ≈ 63.43°

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1+secA/sec A = sin^2 A / 1-cos A​
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Answer:  see proof below

<u>Step-by-step explanation:</u>

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<u>Proof LHS → RHS</u>

\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}

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\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}

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\text{LHS = RHS:}\quad \dfrac{\sin^2 A}{1-\cos A}=\dfrac{\sin^2 A}{1-\cos A}\quad \checkmark

3 0
4 years ago
Need help of 104 please!!!!
Readme [11.4K]

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