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Grade: 12th pass
        
Question: What is the minimum value of sin²θ + cos²θ + sec²θ + cosec²θ + tan²θ + cot²θ?
Answer:  
sin²θ + cos²θ + sec²θ + cosec²θ + tan²θ + cot²θ
= (sin²θ + cosec²θ) + (cos²θ + sec²θ) + (tan²θ + cot²θ)
 
Applying AM-GM logic, [i.e. (a + b) ≥ 2 √a*√b]:
 
Minimum value = (2√ sin²θ *√cosec²θ) + (2√ cos²θ *√ sec²θ) + (2√ tan²θ *√ cot²θ)
                              = (2√1) + (2√1) + (2√1)
                             = 2 + 2 + 2
                                = 6
Is it correct?
9 months ago

Answers : (1)

Utsav Basu
69 Points
							
Sorry Avik, that is wrong.
 
Ans:
sin2θ + cos2θ = 1
sec2θ = 1 + tan2θ
cosec2θ = 1 + cot2θ
 
Using these identities we can simplify the given equation as
   sin²θ + cos²θ + sec²θ + cosec²θ + tan²θ + cot²θ
= (sin²θ + cos²θ) + (sec²θ + tan²θ) + (cosec²θ + cot²θ)
= 1 + 1 + 2tan²θ + 1 + 2cot²θ
= 3 + 2(tan²θ + cot²θ)
 
Now, using AM-GM inequality we can say
min(tan²θ + cot²θ) = 2
 
So, min(3 + 2(tan²θ + cot²θ)) = 3 + 2*2 = 7
9 months ago
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