cos α sin (β – γ) + cos β sin (γ – α) + cos γ sin (α – �

1.	cos α sin (β – γ) + cos β sin (γ – α) + cos γ sin (α – β) is equal to(a) 0 (b) \(\frac12\) (c) 1 (d) 4 cos α cos β cos g
Answer» (a) 0 cos α sin (β – γ) + cos β sin (γ – α) + cos γ sin (α – β) = cos α [sin β cos γ – cos β sin γ] + cos β [sin γ cos α – cos α sin β] + cos γ [sin α cos β – cos α sin β] = cos α sin β cos γ – cos α cos β sin γ + cos β sin γ cos α – cos β cos α sin γ + cos γ sin α cosβ – cos γ cos α sin β = 0.

cos α sin (β – γ) + cos β sin (γ – α) + cos γ sin (α – β) is equal to(a) 0 (b) \(\frac12\) (c) 1 (d) 4 cos α cos β cos g

Answer»

(a) 0

cos α sin (β – γ) + cos β sin (γ – α) + cos γ sin (α – β)

= cos α [sin β cos γ – cos β sin γ] + cos β [sin γ cos α – cos α sin β] + cos γ [sin α cos β – cos α sin β]

= cos α sin β cos γ – cos α cos β sin γ + cos β sin γ cos α – cos β cos α sin γ + cos γ sin α cosβ – cos γ cos α sin β

= 0.