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Find the range of f(x)=cos2x+sec2x.

Answer
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Hint: From the given function using basic identity of (a+b)2, simplify the expression that is given to us. We get f(x)as a function which is greater than zero. Thus form the range of the function.

Complete step by step solution:
The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain. The range is the resulting y-values we get after substituting all the possible x-values. The range of a function is the spread of possible y-values (minimum y-value to maximum y-value)
The function f(x)=cosx has all real numbers in its domain, but its range is 1cosx1. The values of the cosine function are different, depending on whether the angle is in degrees or radians.
Now we have been given the function f(x)=cos2x+sec2x.
Now let us write this above function as,
f(x)=cos2x+sec2x
f(x)=(cosx)2+(secx)2, now apply basic identities to this and simplify the given function and to find the range. We know that,
(a+b)2=a2+2ab+b2a2+b2=(a+b)22aba=cosx,b=secx
Hence our function becomes,
f(x)=(cosx)2+(secx)2=(cosx+secx)22cosxsecx
We know the basic trigonometric function, secx=1cosx. Thus the function becomes,
f(x)=(cosx+secx)2+2cosx×1cosxf(x)=(cosx+secx)2+20
From the above given expression we can see that if we put a number less than 2, we get f(x) as negative. Thus to avoid getting a negative number you can put any values ranging from 2 to infinity. Hence the function obtained is greater than zero and we can write it as,
i.e. f(x)2
Hence we can form its range as,
f(x)=[2,).
Thus we got the required range of the given function f(x)=cos2x+sec2xas,
f(x)=[2,)

Note: You can even substitute f(x)as y, which is much easier to identify. By seeing the question you should be able to identify which property you should be using in order to simplify it. It is important that you should know the range and domain of different functions such as sinx,cosx,tanxetc, so that you might be able to find the range of functions related to it.