If #sin(A), cos(A), tan(A)# are in G.P., then which of the following option is correct #cot^(6)A - cot^(2)A#?

Question

#A) 1#
#B)-1#
#C) 0#
#D)2#

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Answer 1 · 2018-01-11T06:11:29

1

Since #sinA,cosA and tanA# are in GP

#cos^2=sinAtanA#

#=>cos^2=sinAsinA/cosA#

#=>cos^2A/sin^2A=secA#

#=>cot^2A=secA#

#=>cot^4A=sec^2A#

#=>cot^4A=1+tan^2A#

#=>cot^4A=1+1/cot^2A#

#=>cot^6A=cot^2A+1#

#=>cot^6A-cot^2A=1#