Trigonometric functions and their properties

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

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

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

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

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

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

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

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

\(\tan(-x)=\frac{\sin(-x)}{\cos(-x)}=\frac{-\sin x}{\cos x}=-\tan x\); 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 \(f(-x)=-f(x)\).

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 \(60^\circ\).

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

\(\tan\theta=\frac{\sin\theta}{\cos\theta}\). 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 \(\cos\theta=0\).

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 \(90^\circ\).

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.

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

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

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

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

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

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

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

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

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

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