Which of the following trigonometric functions, restricted to 0 < x < π, have values greater than 1?
a.) f(x) = sin x
b.) f(x) = cos x
c.) f(x) = cot x
d.) f(x) = sec x


We will evaluate each function separately in the given interval:
 0 <x <π
 f (x) = sin x
 f (0) = sin 0 = 0
 f (π) = sin π = 0

 f (x) = cos x
 f (0) = cos 0 = 1
 f (π) = cos π = -1

 f (x) = cot x
 f (0) = cot 0 = Cos0 / sin0 = 1/0 = inf
 f (π) = cot π = Cosπ / sinπ = -1 / 0 = -inf
 We can observe that the function f (x) = cot x tends to infinity when it approaches zero.

 f (x) = sec x
 f (π / 2) = sec π / 2 = 1 / cos π / 2 = 1/0 = inf
 We can observe that this function has values that tend towards infinity when it approaches π / 2

 this function has values greater than 1 in 0 <x <π
 f (x) = sec x

cot x and sec x

Step-by-step explanation:

Given are 4 trignometric functions defined in the interval [0,\pi]

We are asked to identity the functions which can take values more than 1.

We have

f(x)=sinx : f(x)= cos x

have values always between -1 and 1 for any value of x

Hence sinx and cosx cannot take values greater than 1.

f(x) =cotx

This value being ratio of cos to sin will become greater than 1 whenever cos >sin value.

SO this function can take values >1


This function being reciprocal of cos function always can take values outside


Hence this value can be greater than 1

 The answer is (c) f(x) = cos x and f(x) = sec x.

Step-by-step explanation: We are given four trigonometric functions out of we are to select those which have values greater than for 0 < x < π.

Since the values of sine and cosine functions lies between -1 and 1. So, they cannot go beyond 1 and hence option (a) and (b) are incorrect.

Since \tan \dfrac{\pi}{6}=\dfrac{1}{\sqrt 3},

so we will have

\cot \dfrac{\pi}{6}=\sqrt {3}1.

So, option (c) is correct.

And, we know that

\cos \dfrac{\pi}{6}=\dfrac{\sqrt 3}{2},

so we will have

\sec \dfrac{\pi}{6}=\dfrac{2}{\sqrt {3}}1.

So, this option is also correct.

Thus, (c) and (d) are the correct ones.

I think the answer is both C and D. A and B are incorrect because sin and cos never go above 1 but both cot and sec go above one within that domain. 

Do you know the answer?

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