It is given that A+B+C=270∘ cos‌2‌A+cos‌2‌B+cos‌2‌C+4‌sin‌A‌sin‌B‌sin‌C =2‌cos(A+B)‌cos(A−B)+1−2sin2C+4‌sin‌A‌sin‌B‌sin‌C =2‌cos(270∘−C)‌cos(A−B)+1−2sin2C+4‌sin‌A‌sin‌B‌sin‌C =1−2‌sin‌C‌cos(A−B)−2sin2C+4‌sin‌A‌sin‌B‌sin‌C =1−2‌sin‌C[cos(A−B)+sin‌C]+4‌sin‌A‌sin‌B‌sin‌C =1−2‌sin‌C[cos(A−B)−cos(A+B)]+4‌sin‌A‌sin‌B‌sin‌C =1−4‌sin‌A‌sin‌B‌sin‌C+4‌sin‌A‌sin‌B‌sin‌C =1