Trigonometric functions and their properties

Correct answer: C. when (\sin \theta=0)

(\cosec \theta=\frac{1}{\sin \theta}), so it is undefined when (\sin \theta=0). A function is not defined when the denominator is zero.

Correct answer: D. when (\sin \theta=0)

(\cot \theta=\frac{\cos \theta}{\sin \theta}), so it is undefined when (\sin \theta=0). In quotient functions, identify the denominator.

Correct answer: C. (\tan \theta)

The period of (\tan \theta) is (\pi), so (\tan(\pi+\theta)=\tan \theta). Knowing periods makes transformations easier.

Correct answer: D. (\sin \theta)

The period of (\sin \theta) is (2\pi), so the value remains (\sin \theta). Adding a full cycle does not change (\sin \theta).

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

The cosine function has period \(2\pi\). Therefore, \(\cos(\theta+2\pi)=\cos\theta\), so \(\cos(2\pi+\theta)=\cos\theta\). The expression \(-\cos\theta\) results when \(\pi\) is added to the angle: \(\cos(\theta+\pi)=-\cos\theta\). Exam tip: Adding or subtracting any integral multiple of \(2\pi\) leaves the values of sine and cosine unchanged.

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

\(\sin\left(\frac{\pi}{2}-\theta\right)=\cos \theta\) is a cofunction relation. In transformations with \(\frac{\pi}{2}\), the function changes.

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

\(\cos\left(\frac{\pi}{2}-\theta\right)=\sin \theta\). In cofunction relations, \(\sin \theta\) and \(\cos \theta\) interchange.

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

\(\tan\left(\frac{\pi}{2}-\theta\right)=\cot \theta\). The cofunction of \(\tan \theta\) is \(\cot \theta\).

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

\(\sec\left(\frac{\pi}{2}-\theta\right)=\cosec \theta\). \(\sec \theta\) and \(\cosec \theta\) are cofunctions.

Correct answer: B. \(\tan x\)

\(\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin x}{\cos x}=-\tan x\); therefore, \(\tan x\) is an odd function. \(\cos x\) and \(\sec x\) are even, and \(\cos^2 x\) is also even. Exam tip: test an odd function using \(f(-x)=-f(x)\).

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

From the standard trigonometric values, \(\cos 60^\circ=\frac{1}{2}\), so option B is correct. \(\frac{\sqrt{3}}{2}\) is the value of \(\cos 30^\circ\), not \(\cos 60^\circ\). Exam tip: Memorise the standard values for \(30^\circ\), \(45^\circ\), and \(60^\circ\).

Correct answer: A. \(\dfrac{1}{\sqrt{3}}\)

Using the standard value, for \(30^\circ\):
\[\tan 30^\circ=\dfrac{1}{\sqrt{3}}.\]

Correct answer: B. \(\frac{\pi}{2}+n\pi\), जहाँ \(n\) पूर्णांक है

\(\tan\theta=\frac{\sin\theta}{\cos\theta}\). It is undefined when its denominator, \(\cos\theta\), is zero. Since \(\cos\theta=0\) at \(\theta=\frac{\pi}{2}+n\pi\), where \(n\) is an integer, option B is correct. At \(n\pi\), \(\tan\theta=0\), and it is defined at \(\frac{\pi}{4}+n\pi\) and \(\frac{\pi}{3}+n\pi\). Exam tip: for tangent, first check where \(\cos\theta=0\).

Correct answer: A. 0

From the standard trigonometric values, \(\cos 90^\circ=0\). On the unit circle, the x-coordinate of the point at \(90^\circ\) is 0, and this coordinate represents cosine. The value 1 is \(\cos 0^\circ\), so it is not correct here. Exam tip: Memorise the standard trigonometric values at \(0^\circ, 30^\circ, 45^\circ, 60^\circ\), and \(90^\circ\).

Correct answer: B. (0)

The correct answer is option B: \\(0\\). Tangent is defined as \\(\tan\theta=\frac{\sin\theta}{\cos\theta}\\). At zero degrees, \\(\sin0^\circ=0\\) and \\(\cos0^\circ=1\\). Hence \\(\tan0^\circ=0/1=0\\). Option A, 1, is not the tangent value; it is associated with some other standard trigonometric values. Option B is correct because the numerator is zero while the denominator is nonzero. Option C, \\(\sqrt3\\), is the tangent of a different standard angle, not 0 degrees. Option D, \\(1/\sqrt3\\), also belongs to another angle and is not applicable here. Do not confuse \\(\tan0^\circ\\) with \\(\tan45^\circ=1\\). Memory cue: at 0 degrees, sine is 0, cosine is 1, so tangent is 0.

Correct answer: C. (1)

(\sec 0^\circ=\frac{1}{\cos 0^\circ}=1). While finding reciprocal values, first write the value of the basic function.

Correct answer: A. (1-\cos^2 \theta)

From (\sin^2 \theta+\cos^2 \theta=1), we get (\sin^2 \theta=1-\cos^2 \theta). In exams, write the basic identity first.

Correct answer: B. (\sec^2 \theta)

(\sec^2 \theta=1+\tan^2 \theta). In questions involving (\tan \theta), recognizing (\sec^2 \theta) is useful.

Correct answer: A. Positive

In the first quadrant, all trigonometric functions are positive. Remember quadrant signs for exams.

Correct answer: B. (\sin \theta)

In the second quadrant, (\sin \theta) and (\cosec \theta) are positive. Knowing sign rules helps solve such questions quickly.

Correct answer: C. Positive

In the third quadrant, (\tan \theta) and (\cot \theta) are positive. The sign changes according to the quadrant.

Correct answer: D. (\cos \theta)

In the fourth quadrant, (\cos \theta) and (\sec \theta) are positive. Remember sign rules using a small table.

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

With \(\frac{\pi}{2}\), \(\sin \theta\) changes to its cofunction \(\cos \theta\). The sign remains positive in the second quadrant.

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

\(\cos\left(\frac{\pi}{2}+\theta\right)=-\sin \theta\). At \(\frac{\pi}{2}\), the function changes and the sign is decided by the quadrant.

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

The cofunction of \(\tan \theta\) is \(\cot \theta\), and the sign is negative in the second quadrant. Therefore the answer is \(-\cot \theta\).