Trigonometric functions and their properties

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

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

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

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

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

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

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

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

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

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

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.

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

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.

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

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

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.

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.

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

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

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.

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.

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.

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

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

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.