Trigonometric functions and their properties

Correct answer: B. (\frac{12}{13})

In ( \cot \theta=\frac{12}{5}), adjacent is (12), opposite is (5), and hypotenuse is (13). In exams use ( \cos \theta=\frac{\text{adjacent}}{\text{hypotenuse}}).

Correct answer: B. (\frac{1}{2})

( \sec \theta=\frac{1}{\cos \theta}), so ( \cos \theta=\frac{1}{2}). In exams remember reciprocal relations.

Correct answer: C. (\frac{2}{5})

( \cosec \theta=\frac{1}{\sin \theta}), so ( \sin \theta=\frac{2}{5}). In exams remember cosecant and sine are reciprocals.

Correct answer: B. (1)

( \cosec \theta=\frac{1}{\sin \theta}), so the product is (1). In exams do not ignore the defined condition.

Correct answer: B. (1)

Since ( \sec \theta=\frac{1}{\cos \theta}), the product becomes (1). In exams identify reciprocal pairs quickly.

Correct answer: C. (1)

( \cot \theta=\frac{1}{\tan \theta}), so the product is (1). In exams reciprocal formulas save time.

Correct answer: B. (0)

( \sin(-\theta)=-\sin \theta), so both terms cancel. In exams odd function property solves such questions quickly.

Correct answer: A. (0)

The direct answer is option A: \(q=0\). To solve this, remember that cosine is an even function. An even function has the same value when its input changes from \(\theta\) to \(-\theta\), so \(\cos(-\theta)=\cos\theta\). Therefore, \(q=\cos\theta-\cos(-\theta)=\cos\theta-\cos\theta=0\). Option A is correct because the two equal quantities cancel. Option B, \(2\cos\theta\), would result from adding the two cosine terms, not subtracting equal terms. Option C, \(-2\cos\theta\), also does not follow from the given subtraction. Option D, 1, is not generally the value of this difference; cosine may have many values depending on \(\theta\). The key exam cue is: cosine is even, so its negative-angle value is unchanged. Do not confuse this with sine, because sine is an odd function and \(\sin(-\theta)=-\sin\theta\).

Correct answer: C. (0)

( \tan(\pi-\theta)=-\tan \theta), so the sum is (0). In exams allied angle sign gives the answer quickly.

Correct answer: B. (-\cos 2\theta)

Because ( \cos 2\theta=\cos^2 \theta-\sin^2 \theta), the given form is (-\cos 2\theta). In exams read double angle identities in reverse too.

Correct answer: B. (\frac{\pi}{4})

In the first quadrant, the maximum of ( \sin \theta+\cos \theta) is ( \sqrt{2}) at ( \theta=\frac{\pi}{4}). In exams use symmetry.

Correct answer: B. \(\sqrt{2}\)

\( \sin \theta+\cos \theta=\sqrt{2}\sin\left(\theta+\frac{\pi}{4}\right)\), so the maximum is \( \sqrt{2}\). In exams use maximum of (a\sin x+b\cos x) as \( \sqrt{a^2+b^2}\).

Correct answer: A. (-\sqrt{2})

The amplitude of ( \sin \theta-\cos \theta) is ( \sqrt{1^2+(-1)^2}=\sqrt{2}). In exams the minimum is the negative amplitude.

Correct answer: A. (5)

The maximum of (a\sin \theta+b\cos \theta) is ( \sqrt{a^2+b^2}), so (M=5). In exams add the squares of coefficients.

Correct answer: B. (-13)

The amplitude is ( \sqrt{5^2+(-12)^2}=13), so the minimum is (-13). In exams maximum is positive amplitude and minimum is negative amplitude.

Correct answer: B. (\sin \theta) negative and (\cos \theta) negative

In the third quadrant, both sine and cosine are negative. In exams use the ASTC rule to decide signs quickly.

Correct answer: A. (\frac{2\pi}{3})

The period of ( \sin k\theta) is ( \frac{2\pi}{k}), so here it is ( \frac{2\pi}{3}). In exams use the coefficient in the period formula.

Correct answer: B. (\theta=(2n+1)\pi)

( \cos \theta=-1) occurs at odd multiples of ( \pi). In exams distinguish the solutions of ( \cos \theta=1) and ( \cos \theta=-1).

Correct answer: A. (\frac{\pi}{2})

The fundamental period of ( \cos k\theta) is ( \frac{2\pi}{k}), so here it is ( \frac{\pi}{2}). In exams put the coefficient in the denominator.

Correct answer: A. (\sec \theta) positive and (\cosec \theta) negative

In the fourth quadrant, (\cos \theta) is positive and (\sin \theta) is negative, so their reciprocals have the same signs. In exams decide the signs of basic functions first.

Correct answer: C. (\tan \theta)

The fundamental period of (\tan \theta) is (\pi), and (5\pi) is a multiple of the period, so the value does not change. In exams simplify tangent by removing multiples of (\pi).

Correct answer: A. 14

Direct answer: 14, so option A is correct. Let x = tan θ and y = cot θ. Since cot θ is the reciprocal of tan θ wherever both are defined, xy = tan θ · cot θ = 1. Use the identity (x + y)² = x² + y² + 2xy. Substituting the given sum, 4² = tan² θ + cot² θ + 2(1). Thus 16 = tan² θ + cot² θ + 2, and the required value is 16 − 2 = 14. Option A wins because it follows directly from the identity. Option B is 16, but it forgets the cross-term 2xy. Option C, 12, does not result from the calculation and has no supporting identity here. Option D, 10, is also an unsupported subtraction. Memory cue: when tan and cot occur together, immediately check tan θ · cot θ = 1, then expand the square carefully.

Correct answer: A. 13

The governing result is that for real constants a and b, the maximum value of a sin x + b cos x is √(a² + b²). This follows from expressing the combination as R sin(x − φ), where R = √(a² + b²), or from the Cauchy–Schwarz inequality. Here a = 5 and b = −12, so R = √(5² + (−12)²) = √(25 + 144) = √169 = 13. Since sine can attain 1 for a suitable phase, this amplitude is attainable and is the maximum. Therefore option A is correct. Option B adds coefficient magnitudes directly, option C subtracts them, and option D uses only one coefficient; none gives the resultant amplitude.