(cos(180−x)sin(x−90)−1)$$÷
tan^2(180+x)\sin(90+x)\cos(-x)$$

Question

(cos(180−x)sin(x−90)−1)$$÷
tan^2(180+x)\sin(90+x)\cos(-x)$$

anonymous · Answer

cot^2(180+x) csc(90+x) sec(x) (-1-cos(180-x) sin(90-x))

anonymous · Answer

thanks for the help but cot and sec is not in our syllabus , we dont learn it in school

anonymous · Answer

oh i see but just to let u know the cot is actually the inverse of tangent and sec is the inverse of cosine......inverse or reciprocal same meaning......so dun worry u just make them 1/(answer of cot) and do that for sec ....ok

anonymous · Answer

okay cool thanks sooo much , i really appreciate it :)

anonymous · Answer

oh sure no worries :) glad to help only if i know the right answer haha...kk