Trigonometric functions and their properties

(\sin x) repeats its value after every (2\pi). In exams first identify the basic function.

The value of (\cos x) always lies between (-1) and (1). The end values are included in the range.

Since (\sin(-x)=-\sin x), (\sin x) is an odd function. For an odd function, (f(-x)=-f(x)).

For an even function, \(f(-x)=f(x)\) for every value in its domain. Since \(\cos(-x)=\cos x\), \(\cos x\) is an even function. In contrast, \(\sin x\), \(\tan x\), and \(\cot x\) are odd functions because they satisfy \(f(-x)=-f(x)\). Exam tip: an even function is symmetric about the \(y\)-axis.

The fundamental Pythagorean trigonometric identity is \(\sin^2 x+\cos^2 x=1\). Therefore, the correct value is 1. Note that \(\sec^2 x=1+\tan^2 x\), so it is not the value of this expression. Exam tip: whenever you see \(\sin^2 x+\cos^2 x\), replace it with 1.

(\tan x) repeats its value after every \(\pi\). The period of (\tan x) and (\cot x) is \(\pi\).

\(\sec x=\frac{1}{\cos x}\), so \(\sec x\) is the reciprocal of \(\cos x\). The reciprocal of \(\tan x\) is \(\cot x\), so option B is not correct. Exam tip: remember the pairs \(\sec\leftrightarrow\cos\), \(\cosec\leftrightarrow\sin\), and \(\cot\leftrightarrow\tan\).

The correct answer is \(\sin x\), because \(\cosec x=\frac{1}{\sin x}\). Thus, \(\cosec x\) is the multiplicative reciprocal of \(\sin x\). \(\sec x\) is the reciprocal of \(\cos x\), so it is not correct here. Exam tip: Remember the reciprocal pairs \(\sin\)–\(\cosec\), \(\cos\)–\(\sec\), and \(\tan\)–\(\cot\).

(\tan x=\frac{\sin x}{\cos x}) is a basic quotient identity. In such questions, check numerator and denominator carefully.

By the quotient identity, \(\cot x=\frac{\cos x}{\sin x}\), so option B is correct. Option A is the formula for \(\tan x=\frac{\sin x}{\cos x}\), not for \(\cot x\). Note that \(\cot x\) is defined only when \(\sin x\ne0\). Exam tip: in \(\tan x\), sine is in the numerator and cosine in the denominator; for \(\cot x\), the order is reversed.

For real values of \(x\), \(\sin x\) always lies between \(-1\) and \(1\). It attains \(1\) at \(x=\frac{\pi}{2}\) and \(-1\) at \(x=\frac{3\pi}{2}\), so its range is \([-1,1]\). The interval \([0,1]\) includes only the non-negative sine values, so it is not the complete range. Exam tip: remember that both sine and cosine have range \([-1,1]\).

\(\tan x=\dfrac{\sin x}{\cos x}\) is defined wherever \(\cos x\ne0\). On every interval where it is defined, \(\tan x\) takes every real value from \(-\infty\) to \(\infty\). Hence its range is \(( -\infty,\infty )\). The interval \([-1,1]\) is the range of \(\sin x\) and \(\cos x\), not of \(\tan x\). Exam tip: the ranges of \(\tan x\) and \(\cot x\) are all real numbers.

Since (\sec^2 x=1+\tan^2 x), (\sec^2 x-\tan^2 x=1). Use the identity in the correct direction.

Since (\cosec^2 x=1+\cot^2 x), the difference is (1). This identity is linked to (\sin^2 x+\cos^2 x=1).

At \(x=0\), \(\cos 0=1\). Therefore, the correct option is \(1\). Note that \(\sin x=0\) can also occur at \(x=n\pi\), but the question specifically asks for \(x=0\). Exam tip: Remember that \(\sin 0=0\) and \(\cos 0=1\).

(\sin 0=0) is a standard value. In exams, remember the values for (0), (\frac{\pi}{2}), and (\pi).

Using \(\tan \theta=\frac{\sin \theta}{\cos \theta}\), we get \(\tan 0=\frac{\sin 0}{\cos 0}=\frac{0}{1}=0\). Hence, 0 is the correct option. It is not undefined because \(\cos 0=1\), not zero. Exam tip: \(\tan \theta\) is undefined only when \(\cos \theta=0\).

At (\frac{\pi}{2}), the value of (\sin x) is (1). The unit circle helps remember this value quickly.

On the unit circle, the point corresponding to the angle \(\frac{\pi}{2}\) is \((0,1)\). Cosine equals the x-coordinate, so \(\cos \frac{\pi}{2}=0\). The value \(-1\) is for \(\cos \pi\), not for \(\cos \frac{\pi}{2}\). Exam tip: Remember that cosine values at \(0, \frac{\pi}{2}, \pi\) are \(1,0,-1\), respectively.

(\pi-x) lies in the second quadrant and (\sin x) remains positive there. Hence (\sin(\pi-x)=\sin x).