Trigonometric functions and their properties

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.

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.

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

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

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

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

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

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

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

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

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

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.

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 \(1\).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

\(\sec x=\frac{1}{\cos x}\), 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 \(\cot\leftrightarrow\tan\).

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 \(\tan\)–\(\cot\).