Trigonometric functions and their properties

\(\sec\left(\frac{\pi}{2}-\theta\right)=\cosec \theta\). \(\sec \theta\) 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 \(60^\circ\).

(\tan 30^\circ=\frac{1}{\sqrt{3}}). Remember (\tan 30^\circ) and (\tan 60^\circ) separately.

\(\tan\theta=\frac{\sin\theta}{\cos\theta}\). 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 \(90^\circ\).

(\tan 0^\circ=0). It can also be checked using (\tan \theta=\frac{\sin \theta}{\cos \theta}).

(\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 \(-\cot \theta\).

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.