If #sinA + cosecA + 2 = 0# then, #sin^(2017)(A) + cosec^(2016)(A) + 1# is equal to?

Question

A) 1
B) 0
C) 2
D) 3

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Answer 1 · 2017-09-13T05:57:59

#sinA+cscA+2=0#

#=>sinA+1/sinA+2=0#

#=>sin^2A+2sinA+1=0#

#=>(sinA+1)^2=0#

#sinA=-1#

#=>cscA=-1#

So

#sin^2017(A)+csc^2016(A)+1#

#=>(-1)^2017+(-1)^2016+1#

#=>-1+1+1=1#