Question

Tan11π/3-2sin4π/6-3/4cosec square π/4+4 cos
square 17π/6

Answer 1

How to solve this: tan−11π3−2sin4π6−34csc2π4+4cos217π6tan⁡−11π3−2sin⁡4π6−34csc2⁡π4+4cos2⁡17π6?

Answer

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1 ANSWER



Suyash Dewangan, Insightful in old fashion ;)

Answered Sep 24, 2015

If I could interpret your question right, it is:
 
tan(11π3)−2sin(4π6)−34cosec2(π4)+4cos2(17π6)tan(11π3)−2sin(4π6)−34cosec2(π4)+4cos2(17π6)
 
= tan(4π−π3)−2sin(2π3)−34cosec2(π4)+4cos2(3π−π6)tan(4π−π3)−2sin(2π3)−34cosec2(π4)+4cos2(3π−π6) 
 
=−tan(π3)−2sin(π−π3)−34cosec2(π4)+4cos2(π6)−tan(π3)−2sin(π−π3)−34cosec2(π4)+4cos2(π6)
 
= −3–√−2sin(π3)−34×(2–√)2+4×(3√2)2−3−2sin(π3)−34×(2)2+4×(32)2
 
= −3–√−2×3√2−34×2+4×34−3−2×32−34×2+4×34
 
=−3–√−3–√−32+3−3−3−32+3
 
= −23–√+32−23+32
 
= 3−43√23−432
 
I hope it helps!