The minimum value of (3sin⁡x−4cos⁡x−10)(3sin⁡x+4cos⁡x−10) is

f(x)=9sin2⁡x−16cos2⁡x−10(3sin⁡x−4cos⁡x)−10(3sin⁡x+4cos⁡x)+100 =25sin2⁡x−60sin⁡x+84=(5sin⁡x−6)2+48The minimum value of f (x) occurs when sin x = 1 .Therefore, the minimum value of f (x) is 49.