Question

Pls solve this question

Answer 1

f(x) = sec2x +  cosec2x  =  1/cos2x   +  1/sin2x  =sin2x + cos2x / sin2x cos2x

Now, we do know the above function would not be defined :

if sin2x = 0   or   cos2x = 0

cause function would approach infinity  and f(x) would discontinuous.

sin2x = 0

2x  = nπ  ⇒ x = nπ/2   where , n = 0, +-1, +-2, +-3 ,...........   or 

cos2x = 0

2x  = (2n +1)π/2   ⇒  x = (2n+1)π /4   where n = 0, +-1 , +-2, +-3, ..........


Union of values of x fetches :  x ∉ nπ  where, n = 0, +-1, +-2 , +-3 ,  ..........

Option A)

If any query , then do ask.

Answer 2

Answer:

nπ/4

Step-by-step explanation:

f(x) = sec2x + cosec2x

f(x) =  (sin2x + cos2x)/(sin2x . cos2x)

f(x) =   (sin2x + cos2x)/(sin2x . cos2x)× 2/2   [multiplying numerator and                    denominator by 2]

f(x)=  2(sin2x + cos2x)/(sin 4x) [2sin2x cos2x=sin4x]

f(x) not to be defined at sin4x = 0 = sin nπ; n∈I

4x = nπ

⇒  x =  ; nπ/4 ∈ I