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

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

6 0

2 years ago

Read 2 more answers

Can someone please help me with this? Thank youuu!

Lunna [17]

Answer:

1996

Step-by-step explanation:

To find the number of terms in an arithmetic sequence, divide the common difference into the difference between the last and first terms, and then add 1.

2008 - 13 = 1995 + 1 = 1996

3 0

2 years ago

What are the coordinates of R’, the image of R(-4,3) after a reflection in the line y=x?

navik [9.2K]

Answer:

R'(3,-4)

Step-by-step explanation:

we know that

When you reflect a point across the line y = x, the x-coordinate and y-coordinate change places

The rule of the reflection across the line y=x is equal to

(x,y) ------> (y,x)

we have

R(-4,3) ------> R'(3,-4)

6 0

2 years ago