<img src="https://tex.z-dn.net/?f=%5Csqrt31" id="TexFormula1" title="\sqrt31" alt="\sqrt31" align="absmiddle" class="latex-formu

$\dfrac{2}{\sqrt3\cos x+\sin x}=\sec\left(\dfrac{\pi}{6}-x\right)\\\\\dfrac{2}{\sqrt3\cos x+\sin x}=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}\ \ \ \ \ (*)\\----------------\\\\\cos\left(\dfrac{\pi}{6}-x\right)\ \ \ \ |\text{use}\ \cos(x-y)=\cos x\cos y+\sin x\sin y\\\\=\cos\dfrac{\pi}{6}\cos x+\sin\dfrac{\pi}{6}\sin x=\dfrac{\sqrt3}{2}\cos x+\dfrac{1}{2}\sin x=\dfrac{\sqrt3\cos x+\sin x}{2}\\------------------------------$

$(*)\\R_s=\sec\left(\dfrac{\pi}{6}-x\right)=\dfrac{1}{\cos\left(\dfrac{\pi}{6}-x\right)}=\dfrac{1}{\dfrac{\sqrt3\cos x+\sin x}{2}}\\\\=\dfrac{2}{\sqrt3\cos x+\sin x}=L_s$

6 0

3 years ago

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Which expression is a difference of squares with a factor of 2x + 5? 4x2 + 10 4x2 – 10 4x2 + 25 4x2 – 25

Shtirlitz [24]

Answer:

4x^2 -25

Step-by-step explanation:

The other factor of the difference of squares is the "conjugate" of the given one: (2x -5). Then their product is ...

(2x +5)(2x -5) = (2x)^2 -(5)^2 = 4x^2 -25

5 0

4 years ago

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