Question

Tan²β+cot²β+2=sec²β*cosec²β (Proved)

Answer 1

Answer:

Sec² β) + (cosec² β)

=( tan²β +1) + (cot²β +1)

= tan²β + cot²β +2

= (sin²β/cos²β) + ( cos²β /sin²β) + 2

= [{sin^4(β) + cos^4(β)}/ {sin²β cos²β}] +2

= { (sin²β + cos²β )² -2sin²β cos²β }/{sin²β * cos²b) +2

= [(1 - 2sin²β cos²β)/(sin²βcos²β)] + 2

Now, for checking whether the value goes less than 2 or not, we replace sinβcosβ by its minimum & maximum value.

minimum value of sinβcosβ =0

maximum value of sinβcosβ = 1/2

=> by putting minimum value , we get

(1–0)/0 +2 , which can not be defined.. & here the first term never becomes negative , to make the over all value less that 2.

Now, by putting maximum value, we get

{(1– 2*1/4)/(1/4)} +2

= {(1 - 1/2)/(1/4)} + 2

= (1/2) x (4) + 2

= 2 + 2 = 4

So, here too, value exceeds .

=> (sec²β + cosec²β ) can exceed 2 but can never be less than 2