#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# 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                                = 6Is it correct?

Utsav Basu
70 Points
3 years ago
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