Trigonometric functions and their properties

Correct answer: C. \(0\)

For every real \(\theta\), \(-1 \leq \cos\theta \leq 1\). Hence, \(\cos^2\theta\) lies in the interval from \(0\) to \(1\). When \(\cos\theta=0\), for example at \(\theta=\frac{\pi}{2}\), we get \(\cos^2\theta=0\). Therefore, the minimum value is \(0\). The value \(1\) is the maximum, not the minimum. Exam tip: the square of a real number cannot be negative.

Correct answer: D. Constant

The governing concept is the fundamental Pythagorean identity of trigonometry: sin² θ + cos² θ = 1 for every real angle θ. The expression therefore always evaluates to the fixed number 1, independent of θ, so it has a constant value. Option D is correct. It is not variable because changing θ changes the individual sine and cosine terms but their squared sum remains unchanged. It is not negative because 1 is positive, and it is not undefined because sine and cosine are defined for every real angle. This identity can also be seen from the unit circle, where the coordinates satisfy x² + y² = 1.

Correct answer: A. (1)

Putting (\sin \theta=0) in (\sin^2 \theta+\cos^2 \theta=1) gives (\cos^2 \theta=1). Use the identity in such questions.

Correct answer: B. (1)

Putting (\cos \theta=0) in (\sin^2 \theta+\cos^2 \theta=1) gives (\sin^2 \theta=1). Substitute the known value in the formula.

Correct answer: C. (\cos^2 \theta)

(\1+\tan^2 \theta=\sec^2 \theta) and (\frac{1}{\sec^2 \theta}=\cos^2 \theta). Be careful while taking reciprocals.

Correct answer: D. (\sin^2 \theta)

(\1+\cot^2 \theta=\cosec^2 \theta) and (\frac{1}{\cosec^2 \theta}=\sin^2 \theta). Simplify fractions using identities.

Correct answer: A. (\tan^2 \theta)

From (\sec^2 \theta=1+\tan^2 \theta), we get (\sec^2 \theta-1=\tan^2 \theta). Rearrange the formula into a simple form.

Correct answer: B. (\cot^2 \theta)

From (\cosec^2 \theta=1+\cot^2 \theta), (\cosec^2 \theta-1=\cot^2 \theta). Practice similar identities in pairs.

Correct answer: C. (\sin^2 \theta)

Since (\cosec \theta=\frac{1}{\sin \theta}), (\frac{\sin \theta}{\cosec \theta}=\sin^2 \theta). Convert reciprocal relations into fractions.

Correct answer: D. (\cos^2 \theta)

Since (\sec \theta=\frac{1}{\cos \theta}), (\frac{\cos \theta}{\sec \theta}=\cos^2 \theta). Writing the reciprocal makes simplification easy.

Correct answer: A. (1)

(\sin \theta) and (\cosec \theta) are reciprocals, so their product is (1). Remember reciprocal pairs.

Correct answer: B. 1

The governing concept is the reciprocal relationship between cosine and secant. By definition, sec θ = 1/cos θ wherever cos θ is nonzero. Hence cos θ · sec θ = cos θ · (1/cos θ) = 1. Therefore option B is correct. This is a reciprocal cancellation, not a squaring operation. Option C would result from multiplying cos θ by itself, while option D would involve multiplying sec θ by itself. Option A is incorrect because a number and its reciprocal do not generally have product zero; their product is one. The usual domain condition is important: the expression is defined only when cos θ ≠ 0.

Correct answer: B. \(1\)

Using the definitions, \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) and \(\cot\theta=\frac{\cos\theta}{\sin\theta}\). Therefore, \(\tan\theta\cdot\cot\theta=\frac{\sin\theta}{\cos\theta}\cdot\frac{\cos\theta}{\sin\theta}=1\), wherever both functions are defined. It is not \(0\); tangent and cotangent are reciprocal ratios. Exam tip: the product of reciprocal trigonometric ratios is usually \(1\).

Correct answer: D. (\sin \theta)

Using (\tan \theta=\frac{\sin \theta}{\cos \theta}) and (\sec \theta=\frac{1}{\cos \theta}), the answer is (\sin \theta). Convert division into fractions and cancel.

Correct answer: A. (\cos \theta)

Using (\cot \theta=\frac{\cos \theta}{\sin \theta}) and (\cosec \theta=\frac{1}{\sin \theta}), we get (\cos \theta). Writing standard forms is the safest method.

Correct answer: B. (\cosec \theta)

From (\sec \theta=\frac{1}{\cos \theta}) and (\tan \theta=\frac{\sin \theta}{\cos \theta}), we get (\frac{1}{\sin \theta}=\cosec \theta). Simplify fractions carefully.

Correct answer: A. \(\sec \theta\)

Using \(\cosec\theta=\frac{1}{\sin\theta}\) and \(\cot\theta=\frac{\cos\theta}{\sin\theta}\),
\(\frac{\cosec\theta}{\cot\theta}=\frac{1/\sin\theta}{\cos\theta/\sin\theta}=\frac{1}{\cos\theta}=\sec\theta\). Hence, \(\sec\theta\) is correct. \(\tan\theta\) equals \(\frac{\sin\theta}{\cos\theta}\), so it is not the answer. Exam tip: rewrite reciprocal and quotient trigonometric ratios in terms of \(\sin\theta\) and \(\cos\theta\) before simplifying.

Correct answer: C. 2π

The governing concept is the period of a periodic function: the least positive number T such that f(x + T) = f(x) for every x. For the sine function, sin(x + 2π) = sin x, so its fundamental period is 2π radians. This represents one complete revolution on the unit circle, after which the same y-coordinate pattern repeats. Therefore option C is correct. Although 4π also returns the same value, it is not the least positive period. The value π is only half a revolution and gives sin(x + π) = −sin x, so it changes the sign rather than reproducing the function. Similarly, π/2 is a quarter-cycle and cannot reproduce every sine value.

Correct answer: B. [-1, 1]

The range of a function is the set of all output values it can take. On the unit circle, cos x represents the horizontal coordinate, whose value can never be less than −1 or greater than 1. Equivalently, the basic cosine graph reaches a maximum of 1 and a minimum of −1, and both values occur, for example cos 0 = 1 and cos π = −1. Hence the range is the closed interval [−1, 1], so option B is correct. Option A excludes negative cosine values, option C is far too broad, and option D excludes all values below 1. The brackets show that the endpoints are included.

Correct answer: C. sin x

The governing definition is that a function is odd when f(−x) = −f(x) for every x in its symmetric domain. For sine, sin(−x) = −sin x, so sin x is odd and option C is correct. The distractors are even functions: cos(−x) = cos x, so cos x is even. Since sec x = 1/cos x, wherever it is defined, sec(−x) = sec x, making sec x even as well. Also, cos²(−x) = [cos(−x)]² = cos²x, so cos²x is even. Thus “odd” refers to origin symmetry and the sign-change rule, not merely to whether an expression appears to contain an odd power.

Correct answer: D. \(\cos x\)

For an even function, \(f(-x)=f(x)\) for every value in its domain. Since \(\cos(-x)=\cos x\), \(\cos x\) is an even function. In contrast, \(\sin x\), \(\tan x\), and \(\cot x\) are odd functions because they satisfy \(f(-x)=-f(x)\). Exam tip: an even function is symmetric about the \(y\)-axis.

Correct answer: B. 1

The fundamental Pythagorean trigonometric identity is \(\sin^2 x+\cos^2 x=1\). Therefore, the correct value is 1. Note that \(\sec^2 x=1+\tan^2 x\), so it is not the value of this expression. Exam tip: whenever you see \(\sin^2 x+\cos^2 x\), replace it with 1.

Correct answer: B. \(\pi\)

For tangent, \(\tan(x+\pi)=\tan x\) for all \(x\) where both sides are defined. Hence the smallest positive period is \(\pi\).

Correct answer: C. \(\cos x\)

\(\sec x=\frac{1}{\cos x}\), so \(\sec x\) is the reciprocal of \(\cos x\). The reciprocal of \(\tan x\) is \(\cot x\), so option B is not correct. Exam tip: remember the pairs \(\sec\leftrightarrow\cos\), \(\cosec\leftrightarrow\sin\), and \(\cot\leftrightarrow\tan\).

Correct answer: B. \(\sin x\)

The correct answer is \(\sin x\), because \(\cosec x=\frac{1}{\sin x}\). Thus, \(\cosec x\) is the multiplicative reciprocal of \(\sin x\). \(\sec x\) is the reciprocal of \(\cos x\), so it is not correct here. Exam tip: Remember the reciprocal pairs \(\sin\)–\(\cosec\), \(\cos\)–\(\sec\), and \(\tan\)–\(\cot\).