Question

If x/costheta = y/cos(theta-2π/3) = z/cos(theta+2π/3),then x+y+z = ??​

Answer 1

Note : Theta is written as A.

Answer:

Required numeric value of x + y + z is 0.

Step-by-step explanation:

Given,

$\dfrac{x}{cosA}=\dfrac{y}{cos\bigg(A-\dfrac{2\pi}{3} \bigg)} = \dfrac{z}{cos\bigg(A+\dfrac{2\pi}{3}\bigg)}$

Let,

$\dfrac{x}{cosA}=\dfrac{y}{cos\bigg(A-\dfrac{2\pi}{3} \bigg)} = \dfrac{z}{cos\bigg(A+\dfrac{2\pi}{3}\bigg)}=k$

Thus,

• x = k cosA
• y = k cos$\bigg(A-\dfrac{2\pi}{3}\bigg)$
• z = k cos$\bigg(A+\dfrac{2\pi}{3}\bigg)$

Therefore,

= > x + y + z

= > k cosA + k cos$\bigg(A-\dfrac{2\pi}{3}\bigg)$ + k cos$\bigg(A+\dfrac{2\pi}{3}\bigg)$

= > k$\bigg[$cosA + cos$\bigg(A-\dfrac{2\pi}{3}\bigg)$ + cos$\bigg(A+\dfrac{2\pi}{3}\bigg)\bigg]$

From the properties of trigonometry :

• cos( A + B ) = cosAcosB + sinAsinB
• cos( A - B ) = cosAcosB - sinAsinB
• cos( A + B ) + cos( A - B ) = 2cosAcosB

Then, { continued }

= > k$\bigg[$cosA + 2cosAcos$\dfrac{2\pi}{3}\bigg]$

= > k[ cosA + 2cosA( - 1 / 2 ) ] { cos2π/3 = - 1 / 2 }

= > k[ cosA + cosA( - 1 / 2 x 2 ) ] { 2cosB = cos2B }

= > k[ cosA + cos( - A ) ]

= > k[ cosA - cosA ] { cos( - B ) = - cosB }

= > k( 0 )

= > 0

Hence the required numeric value of x + y + z is 0.