3 tan^6 10 - 27 tan^4 10 + 33 tan^2 10 is equal to-    (needs complete solution)

We know that tan3A=3tanA-tan^3 A/1-3tan^2 A
substitute A=10
tan3(10)=3tan10-tan^3 10/1-3tan^2 10
1/31/2=3tan10-tan^3/1-3tan^2 10
1-3tan^2 10=31/2(3 tan10-tan^3 10)
squaring on both sides
1+9tan^4 10-6tan^2 10=27tan^2 10+3tan^6 10-18tan^4 10
3tan^6-27tan^4 10+33 tan^2 10=1
         
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