The correct option is B f(x)=sgn(x2−6x+10),g(x)=sgn(cos2x+sin2(x+π3)) where sgn denotes signum function.
Functions f(x) and g(x) are said to be equal or identical function's if
(i) Domain of f(x)=g(x) and
(ii) Range of f(x)=g(x)
For f(x)=secxcosx−tanxcotx,g(x)=cosxsecx+sinxcosec x
Domain of both functions is x∈R−{nπ2,n∈Z}
Also, both functions simplify to 1 i.e. Range ={1}
Hence, both functions are identical.
For f(x)=sgn(x2−6x+10),g(x)=sgn(cos2x+sin2(x+π3))
As x2−6x+10=(x−3)2+1>0,cos2x+sin2(x+π3)>0
Hence f(x)=g(x)=1 ∀ x∈R. and Domain of f(x)=g(x)=R
Hence both functions are identical.
For f(x)=eln(x2−5x+6),g(x)=x2−5x+6
Domain of f(x) is x∈(−∞,2)∪(3,∞) but domain of g(x)=R
Hence, they are not identical functions.
For
f(x)=sinxsecx+cosxcosec x,g(x)=2cos2xcotx
f(x)=sinxsecx+cosxcosec x
=2sinxcosx
=2cos2xcotx
=g(x)
And domain of both the functions is also same.
Hence, they are identical functions.