Question

class 11th ,please solve question urgent....

Answer 1

cosecx+cotx=√3
now we know that
cosec²x-cot²x=1
(cosecx+cotx)(cosecx-cotx)=1..........(a+b)(a-b)=a²-b²
now
√3(cosecx-cotx)=1
(cosecx-cotx)=1/√3
solving further
cosecx+cotx=√3
cosecx-cotx=1/√3
adding both equations we get
2cosecx=√3+(1/√3)
2cosecx=(3+1)/√3
2cosecx=4/√3
cosecx=2/√3
we know that at x=60° cosecx=2/√3
therefore x= 60°

sorry for any typing error

Answer 2

26.cotx+cosecx =√3
⇒cosxsinx+1sinx=√3
⇒cosx+1 =√3sinx
squaring both sides we get
⇒cos2x+1+2cosx =3sin2x
⇒cos2x+1+2cosx =3(1−cos2x) (as cos2x+sin2x=1)
⇒cos2x+1+2cosx = 3−3cos2x
⇒ 4cos2x+2cosx−2 =0
⇒ 4cos2x+4cosx−2cosx−2=0
⇒4cosx(cosx+1)−2(cosx+1)=0
⇒(4cosx−2)(cosx+1)=0(4cosx−2) =0
⇒cosx =1/2 
cosx is +ve in first and fourth quadrantso
 x = π/3 and x =(2π−π3)=5π3and 
(cosx+1) =0 or cosx = −1
So x =π
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thank u