Trigonometric functions and their properties

Correct answer: D. -\(\tan \theta\)

The cofunction of \(\cot \theta\) is \(\tan \theta\), and the sign is negative in the second quadrant. In such questions, first identify the cofunction.

Correct answer: A. -\(\cosec \theta\)

The cofunction of \(\sec \theta\) is \(\cosec \theta\), and \(\sec \theta\) is negative in the second quadrant. Hence the value is \(-\cosec \theta\).

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

The cofunction of \(\cosec \theta\) is \(\sec \theta\), and the sign remains positive in the second quadrant. Remember cofunction pairs.

Correct answer: C. -\(\cos \theta\)

With \(\frac{3\pi}{2}\), \(\sin \theta\) changes to a cofunction and the sign is negative in the third quadrant. So the answer is \(-\cos \theta\).

Correct answer: A. -\(\sin \theta\)

With \(\frac{3\pi}{2}\), \(\cos \theta\) changes to the cofunction \(\sin \theta\). In the third quadrant, \(\cos \theta\) is negative.

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

With \(\frac{3\pi}{2}\), \(\tan \theta\) changes to \(\cot \theta\). In the third quadrant, \(\tan \theta\) is positive.

Correct answer: C. -\(\cos \theta\)

\(\sin\left(\frac{3\pi}{2}+\theta\right)=-\cos \theta\). In a \(\frac{3\pi}{2}\) form, \(\sin \theta\) changes to its cofunction.

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

\(\cos\left(\frac{3\pi}{2}+\theta\right)=\sin \theta\). In the fourth quadrant, \(\cos \theta\) is taken as positive.

Correct answer: A. -\(\cot \theta\)

With \(\frac{3\pi}{2}\), \(\tan \theta\) changes to \(\cot \theta\), and the sign is negative in the fourth quadrant. Therefore the answer is \(-\cot \theta\).

Correct answer: B. (\sin \theta)

The period of (\sin \theta) is (2\pi), so adding (2\pi) does not change its value. Use periods to simplify large angles.

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

The cosine function has period \(2\pi\). Therefore, adding \(2\pi\) returns the point to the same position on the unit circle, giving \(\cos(\theta+2\pi)=\cos\theta\). In contrast, adding \(\pi\) gives \(\cos(\theta+\pi)=-\cos\theta\). Exam tip: for both sine and cosine, adding \(2\pi\) leaves the value unchanged.

Correct answer: D. (\tan \theta)

The period of (\tan \theta) is (\pi), so (\tan(\theta+\pi)=\tan \theta). For (\tan \theta), (2\pi) is not needed.

Correct answer: A. cot θ

The governing concept is the periodicity of the cotangent function. Since cot x = cos x/sin x, adding π changes the signs of both numerator and denominator: cos(x + π) = −cos x and sin(x + π) = −sin x. Therefore cot(x + π) = [−cos x]/[−sin x] = cos x/sin x = cot x, wherever cotangent is defined. With x = θ, cot(θ + π) = cot θ. Hence option A is correct. Options B and D incorrectly assume that adding π changes the overall sign, forgetting that both signs cancel in the quotient. Option C confuses cotangent with its reciprocal, tangent. The result expresses the period π of cotangent.

Correct answer: B. (\sec \theta)

The period of (\sec \theta) is (2\pi) because it is the reciprocal of (\cos \theta). Hence adding (2\pi) keeps the value same.

Correct answer: C. (\cosec \theta)

The period of (\cosec \theta) is (2\pi) because it is the reciprocal of (\sin \theta). The period gives the answer directly.

Correct answer: D. −sin θ

The governing concept is the sine addition identity and the effect of a half-turn. Using sin(a + b) = sin a cos b + cos a sin b with a = θ and b = π gives sin(θ + π) = sin θ cos π + cos θ sin π. Since cos π = −1 and sin π = 0, this becomes sin θ(−1) + cos θ(0) = −sin θ. Therefore option D is correct. Geometrically, adding π radians, or 180°, moves a point on the unit circle to the opposite point. The y-coordinate, represented by sine, changes sign. Option A ignores that reversal, while options B and C replace sine with cosine and therefore use the wrong trigonometric function.

Correct answer: A. −cos θ

The governing concept is the cosine addition identity. Apply cos(a + b) = cos a cos b − sin a sin b with a = θ and b = π. Since cos π = −1 and sin π = 0, we obtain cos(θ + π) = cos θ(−1) − sin θ(0) = −cos θ. Hence option A is correct. The same result is clear from the unit circle: adding π radians rotates a point by 180°, changing both coordinates to their opposites. Cosine represents the x-coordinate, so it becomes negative. Option B incorrectly assumes no change after the rotation. Options C and D confuse cosine with sine, and D also assigns the wrong sign and function. Thus the identity is valid for every θ for which the expression is considered.

Correct answer: C. -\(\cos \theta\)

\(\cos(\theta-\pi)=-\cos \theta\). A difference of \(\pi\) changes the sign of \(\cos \theta\).

Correct answer: D. (\tan \theta)

The period of (\tan \theta) is (\pi), so (\tan(\theta-\pi)=\tan \theta). Subtracting the period keeps the value same.

Correct answer: A. −sin θ

The governing concept is the sine reflection identity. Use sin(2π − θ) = sin 2π cos θ − cos 2π sin θ, which follows from the sine difference formula. Since sin 2π = 0 and cos 2π = 1, the expression becomes 0·cos θ − 1·sin θ = −sin θ. Therefore option A is correct. On the unit circle, the angle 2π − θ lies in the fourth quadrant when θ is in the first quadrant; its y-coordinate is the negative of the y-coordinate at θ. Option B has the wrong sign, while C and D incorrectly replace sine by cosine.

Correct answer: B. cos θ

Apply the cosine difference identity: cos(a − b) = cos a cos b + sin a sin b. For a = 2π and b = θ, cos(2π − θ) = cos 2π cos θ + sin 2π sin θ. The standard values cos 2π = 1 and sin 2π = 0 give 1·cos θ + 0·sin θ = cos θ. Hence option B is correct. Geometrically, reflection of the angle θ across the x-axis preserves the x-coordinate on the unit circle, and cosine is that x-coordinate. Option A incorrectly changes its sign; options C and D confuse the x-coordinate function with the y-coordinate function.

Correct answer: C. −tan θ

The governing concept is the tangent difference identity, or the quotient tan x = sin x / cos x. From the related transformations, sin(2π − θ) = −sin θ and cos(2π − θ) = cos θ. Therefore tan(2π − θ) = sin(2π − θ) / cos(2π − θ) = (−sin θ)/(cos θ) = −tan θ, provided the expressions are defined. Thus option C is correct. The same result follows because tangent is negative in the fourth quadrant for a first-quadrant reference angle. Options A and B omit or change the function incorrectly, while D has both the wrong reciprocal and the wrong sign.

Correct answer: D. (\sec \theta)

(\sec \theta) is the reciprocal of (\cos \theta) and remains positive in the fourth quadrant. So (\sec(2\pi-\theta)=\sec \theta).

Correct answer: A. -(\cosec \theta)

(\cosec \theta) is the reciprocal of (\sin \theta) and is negative in the fourth quadrant. Therefore the value is (-\cosec \theta).

Correct answer: B. (1)

The value of (\sin \theta) lies in ([-1,1]), so the maximum value of (\sin^2 \theta) is (1). Squaring makes the value non-negative.