Trigonometric functions and their properties

(\tan x=\frac{\sin x}{\cos x}) is a basic quotient identity. In such questions, check numerator and denominator carefully.

By the quotient identity, \(\cot x=\frac{\cos x}{\sin x}\), so option B is correct. Option A is the formula for \(\tan x=\frac{\sin x}{\cos x}\), not for \(\cot x\). Note that \(\cot x\) is defined only when \(\sin x\ne0\). Exam tip: in \(\tan x\), sine is in the numerator and cosine in the denominator; for \(\cot x\), the order is reversed.

For real values of \(x\), \(\sin x\) always lies between \(-1\) and \(1\). It attains \(1\) at \(x=\frac{\pi}{2}\) and \(-1\) at \(x=\frac{3\pi}{2}\), so its range is \([-1,1]\). The interval \([0,1]\) includes only the non-negative sine values, so it is not the complete range. Exam tip: remember that both sine and cosine have range \([-1,1]\).

\(\tan x=\dfrac{\sin x}{\cos x}\) is defined wherever \(\cos x\ne0\). On every interval where it is defined, \(\tan x\) takes every real value from \(-\infty\) to \(\infty\). Hence its range is \(( -\infty,\infty )\). The interval \([-1,1]\) is the range of \(\sin x\) and \(\cos x\), not of \(\tan x\). Exam tip: the ranges of \(\tan x\) and \(\cot x\) are all real numbers.

The direct answer is option D, \(1\). A basic identity connects secant and tangent: \(\sec^2x=1+\tan^2x\), wherever these functions are defined. Subtract \(\tan^2x\) from both sides: \(\sec^2x- an^2x=1\). Option A, 0, would incorrectly treat the two terms as equal. Option B, \(\sin x\), is unrelated to this fixed identity and is not constant. Option C, 2, results from an invalid addition or sign mistake. Option D is correct because the difference is exactly the constant 1. This identity follows from \(\sec x=1/\cos x\), \(\tan x=\sin x/\cos x\), and \(\sin^2x+\cos^2x=1\). Exam cue: remember \(\sec^2x-\tan^2x=1\).

The governing identity is cosec²x = 1 + cot²x. It follows from sin²x + cos²x = 1 by dividing every term by sin²x, wherever the expressions are defined. Subtracting cot²x from both sides gives cosec²x − cot²x = 1. Therefore option B is correct. Options A and D may simplify to 1 when their denominators are nonzero, but they are unrelated forms rather than the identity being directly applied, and they can be undefined for particular x-values. Option C is incorrect because cosec²x and cot²x are not generally equal; their difference is exactly the constant 1. The identity is valid on the common domain of the original expressions.

At \(x=0\), \(\cos 0=1\). Therefore, the correct option is \(1\). Note that \(\sin x=0\) can also occur at \(x=n\pi\), but the question specifically asks for \(x=0\). Exam tip: Remember that \(\sin 0=0\) and \(\cos 0=1\).

The governing concept is the standard value of sine at zero radians. On the unit circle, angle 0 corresponds to the point (1, 0), and the sine of an angle is the y-coordinate of that point. Therefore sin 0 = 0, so option B is correct. This also agrees with the right-triangle interpretation as the angle approaches zero: the opposite side becomes zero relative to the hypotenuse. Option A is cos 0, not sin 0. Option C is associated with sin(3π/2), and option D is a standard value for some other special angles, such as π/6, but not for zero. The angle is understood in radians, although zero has the same value in any angular unit.

Using \(\tan \theta=\frac{\sin \theta}{\cos \theta}\), we get \(\tan 0=\frac{\sin 0}{\cos 0}=\frac{0}{1}=0\). Hence, 0 is the correct option. It is not undefined because \(\cos 0=1\), not zero. Exam tip: \(\tan \theta\) is undefined only when \(\cos \theta=0\).

The governing concept is the standard sine value at a right angle. The angle \(\pi/2\) radians is equal to \(90^\circ\). On the unit circle, the point reached at this angle is \((0,1)\), and sine is the point’s y-coordinate. Consequently, \(\sin(\pi/2)=1\), so option B is correct. The same result follows from a right triangle: at the limiting right-angle position, the opposite-to-hypotenuse ratio is 1. Option A is \(\sin 0\) and \(\sin\pi\), option C is \(\sin(3\pi/2)\), and option D is associated with angles such as \(\pi/6\), not \(\pi/2\).

On the unit circle, the point corresponding to the angle \(\frac{\pi}{2}\) is \((0,1)\). Cosine equals the x-coordinate, so \(\cos \frac{\pi}{2}=0\). The value \(-1\) is for \(\cos \pi\), not for \(\cos \frac{\pi}{2}\). Exam tip: Remember that cosine values at \(0, \frac{\pi}{2}, \pi\) are \(1,0,-1\), respectively.

(\pi-x) lies in the second quadrant and (\sin x) remains positive there. Hence (\sin(\pi-x)=\sin x).

Using the supplementary-angle identity, \(\cos(\pi-x)=-\cos x\). The cosine of \(\pi-x\) has the same magnitude as \(\cos x\), but the opposite sign. Note that \(\sin(\pi-x)=\sin x\), so the sine options do not apply here. Exam tip: for \(\pi-x\), remember that sine remains positive while cosine changes sign.

The correct answer is option C: \\(\tan x\\). Tangent has period \\(\pi\\) radians, meaning adding \\(\pi\\) does not change its value: \\(\tan(x+\pi)=\tan x\\), wherever tangent is defined. This can also be seen from sine and cosine: both \\(\sin(x+\pi)=-\sin x\\) and \\(\cos(x+\pi)=-\cos x\\), so their ratio is \\(\frac{-\sin x}{-\cos x}=\frac{\sin x}{\cos x}=\tan x\\). Option A, \\(-\tan x\\), has the wrong sign. Option B, \\(\cot x\\), is the reciprocal of tangent, not the same function. Option D, \\(-\cot x\\), is also a different function. Option C matches the period identity. Exam cue: for tangent, adding or subtracting \\(\pi\\) leaves the value unchanged.

Sine is an odd function. Therefore, changing the sign of the angle changes the sign of its value: \(\sin(-x)=-\sin x\). Choosing \(\sin x\) would apply the even-function rule, so it is incorrect here. Exam tip: \(\sin\) and \(\tan\) are odd functions, whereas \(\cos\) is an even function.

The governing concept is the even-function property of cosine. A function is even when \(f(-x)=f(x)\); cosine has this property because reflection of an angle across the x-axis leaves the x-coordinate on the unit circle unchanged. Therefore, \(\cos(-x)=\cos x\), making option C correct. This should not be confused with sine, which is odd and satisfies \(\sin(-x)=-\sin x\). Hence options A and D introduce an incorrect negative sign or an incorrect function, while option B changes cosine into sine without justification. The identity is valid for every x for which cosine is defined.

(\tan x) is an odd function, so (\tan(-x)=-\tan x). This can also be understood from (\frac{\sin x}{\cos x}).

In the second quadrant, only (\sin x) and (\cosec x) are positive. Therefore, (\cos x) is negative.

The correct answer is option C: positive. In the third quadrant, both sine and cosine are negative. Since \\(\tan x=\frac{\sin x}{\cos x}\\), the quotient of two negative numbers is positive. Therefore tangent is positive in the third quadrant. Option A, negative, is wrong because it would result when sine and cosine have opposite signs, as in the second or fourth quadrant. Option B, zero, occurs on axes at angles where sine is zero, not throughout the third quadrant. Option C is correct because negative divided by negative is positive. Option D, undefined, occurs where \\(\cos x=0\\), mainly on the vertical axis, not for ordinary angles inside the third quadrant. The sign pattern is remembered by ASTC: All, Sine, Tangent, Cosine; tangent is positive in quadrant III.

In the fourth quadrant, the y-coordinate of a point is negative. Since \(\sin x=\frac{y}{r}\) and \(r\) is always positive, \(\sin x\) is negative. It is zero only on the x-axis, not within a quadrant. Exam tip: In the fourth quadrant, remember that cosine is positive and sine is negative.

The sine function has period \(2\pi\). Hence, adding \(2\pi\) to any angle \(x\) does not change its sine: \(\sin(2\pi+x)=\sin x\). The value \(-\sin x\) occurs on adding \(\pi\), since \(\sin(\pi+x)=-\sin x\). Exam tip: remember \(\sin(x+2n\pi)=\sin x\).

The cosine function has period \(2\pi\). Therefore, adding \(2\pi\) to any angle \(x\) does not change its cosine: \(\cos(x+2\pi)=\cos x\). Hence, \(\cos x\) is correct. The expression \(-\cos x\) occurs on adding \(\pi\), since \(\cos(x+\pi)=-\cos x\). Exam tip: Remember that both \(\sin x\) and \(\cos x\) have period \(2\pi\).

The direct answer is option C, \(\cos x\). The angles \(x\) and \(\frac{\pi}{2}-x\) are complementary. In a right triangle, the sine of one acute angle equals the cosine of its complementary angle, so \(\sin(\frac{\pi}{2}-x)=\cos x\). Option A, \(\sin x\), incorrectly leaves the function unchanged. Option B, \(-\sin x\), introduces a negative sign that is not caused by this complementary-angle change. Option C is correct. Option D, \(-\cos x\), has the correct function but an unnecessary negative sign. The identity can also be checked from the unit-circle or angle-subtraction formula: \(\sin(\frac\pi2-x)=\sin\frac\pi2\cos x-\cos\frac\pi2\sin x=\cos x\). Memory cue: sine and cosine exchange when an angle is subtracted from \(\frac\pi2\).

By the complementary angle identity, (\cos(\frac{\pi}{2}-x)=\sin x). In such questions, (\sin x) and (\cos x) interchange.

The governing concept is the complementary-angle, or cofunction, identity. Express tangent as a quotient of sine and cosine: tan(π/2 − x) = sin(π/2 − x) / cos(π/2 − x). The complementary-angle identities give sin(π/2 − x) = cos x and cos(π/2 − x) = sin x. Therefore tan(π/2 − x) = cos x / sin x = cot x, wherever the expressions are defined. Thus option B is correct. Option A incorrectly treats tangent as unchanged, while options C and D introduce a negative sign that does not arise in this complementary-angle transformation. A negative sign would be associated with a different identity, such as tan(π − x) = −tan x.