Trigonometric functions and their properties

From (\cosec^2 x=1+\cot^2 x), (|\cosec x|=\frac{5}{4}). In the second quadrant, (\cosec x) is positive.

Using the identity \(\sec^2 x=1+\tan^2 x\), we get \(\sec^2 x-1=\tan^2 x\). Therefore, \(\frac{\sec^2 x-1}{\tan^2 x}=\frac{\tan^2 x}{\tan^2 x}=1\), wherever the given expression is defined. Option \(0\) is incorrect because the numerator and denominator are equal; when they are zero, the fraction is undefined rather than zero. Exam tip: Replace \(\sec^2 x-1\) directly with \(\tan^2 x\).

Using the Pythagorean trigonometric identity \(\cosec^2 x=1+\cot^2 x\), we get \(\cosec^2 x-1=\cot^2 x\). Therefore, \(\frac{\cosec^2 x-1}{\cot^2 x}=\frac{\cot^2 x}{\cot^2 x}=1\), wherever the expression is defined. \(\cot^2 x\) is not the final answer because it is only the simplified numerator. Exam tip: remember \(\cosec^2 x=1+\cot^2 x\) alongside \(\sec^2 x=1+\tan^2 x\).

Since \(\cos(-x)=\cos x\), \(\cos x\) is an even function. In contrast, \(\sin(-x)=-\sin x\), so \(\sin x\), \(\tan x\), and \(\cot x\) are odd. Exam tip: test parity by replacing \(x\) with \(-x\).

Using the subtraction formula, \(\sin(\pi-x)=\sin\pi\cos x-\cos\pi\sin x=0-(-1)\sin x=\sin x\). The angle \(\pi-x\) is in quadrant II, where sine is positive. Exam tip: check the quadrant sign for identities involving \(\pi\pm x\).

\\(\tan x=\frac{\sin x}{\cos x}\\). At \\(x=\frac{\pi}{2}+n\pi\\), \\(\cos x=0\\), so division by zero makes \\(\tan x\\) undefined. Exam tip: for a quotient function, first check where its denominator is zero.

The value of \(\sin x\) always lies from \(-1\) to \(1\), and both endpoints occur: \(\sin(\pi/2)=1\) and \(\sin(3\pi/2)=-1\). Hence the endpoints must be included. Exam tip: use inclusive inequality signs for the range of sine.

(\cos^2 x=1-\frac{1}{4}=\frac{3}{4}). In the first quadrant, (\cos x=\frac{\sqrt{3}}{2}) is taken.

The basic cosine function \(\cos t\) repeats after an increase of \(2\pi\) in its angle. For a function of the form \(\cos(kx)\), the input angle changes by \(2\pi\) when \(kx\) changes by \(2\pi\). Thus \(kT=2\pi\), so \(T=2\pi/k\). Here \(k=5\), giving \(T=2\pi/5\). Therefore option A is correct. The factor 5 compresses the graph horizontally and makes the period five times smaller than the usual cosine period. Option D is the period of \(\cos x\), option B is half the required period, and option C incorrectly multiplies rather than divides by the coefficient.

For a sine function of the form \(a\sin bx\), the amplitude is \(|a|\). Here, \(a=4\), so the amplitude is \(|4|=4\). The factor \(2\) affects the period of the function, not its amplitude. Exam tip: To find amplitude, take the absolute value of the coefficient outside \(\sin\) or \(\cos\).

Since \(\cos x\) lies between \(-1\) and \(1\), \(-3\cos x\) lies between \(-3\) and \(3\). When \(\cos x=1\), we get \(-3\cos x=-3\); hence the minimum value is \(-3\). The value \(3\) is the maximum, obtained when \(\cos x=-1\). Exam tip: multiplying a trigonometric function by a negative number reverses the order of its maximum and minimum values.

Since \(\sin x\) always lies between \(-1\) and \(1\), the minimum value of \(2+\sin x\) is \(2-1=1\) and its maximum value is \(2+1=3\). Hence, its range is \([1,3]\). The interval \([-1,1]\) is the range of \(\sin x\) alone, not of \(2+\sin x\). Exam tip: when a constant is added to a function, add that constant to both endpoints of its range.

Since \(\cos x\) lies in \([-1,1]\), the minimum value of \(1-\cos x\) is \(0\) when \(\cos x=1\), and its maximum value is \(2\) when \(\cos x=-1\). Hence, the range is \([0,2]\). The interval \([-1,1]\) is the range of \(\cos x\), not of \(1-\cos x\). Exam tip: for an expression such as \(a-\cos x\), substitute the endpoint values \(-1\) and \(1\) of \(\cos x\).

Dividing numerator and denominator by (\cos x) gives (\frac{\tan x+1}{\tan x-1}). Substituting (\tan x=3) gives (2).

Substitute the given trigonometric values directly into the expression. It becomes \\(\\frac{2+\\sqrt3}{2-\\sqrt3}\\). To simplify it, multiply numerator and denominator by the conjugate of the denominator, \\(2+\\sqrt3\\). The denominator becomes \\(2^2-(\\sqrt3)^2=4-3=1\\), while the numerator becomes \\( (2+\\sqrt3)^2=4+4\\sqrt3+3=7+4\\sqrt3\\).

Therefore the value of the expression is \\(7+4\\sqrt3\\), which is option A. The denominator is positive because 2 is greater than \\(\\sqrt3\\), so no sign issue arises. Rationalising is useful here because it removes the radical from the denominator and gives the exact simplified value.

\(\tan x=\frac{\sin x}{\cos x}\). For \(x=\frac{(2n+1)\pi}{2}\), \(\cos x=0\), so \(\tan x\) is undefined. In contrast, \(\sin x\) and \(\cos x\) are defined for every real \(x\). Exam tip: for a quotient function, first check where its denominator becomes zero.

Using the identity \(\cosec^2 x-\cot^2 x=1\), we get \((\cosec x-\cot x)(\cosec x+\cot x)=1\). Given \(\cosec x-\cot x=\frac{1}{3}\), so \(\frac{1}{3}(\cosec x+\cot x)=1\). Therefore, \(\cosec x+\cot x=3\). The value \(\frac{1}{3}\) is the given expression; its reciprocal gives 3. Exam tip: recognise \(\cosec^2 x-\cot^2 x=1\) and multiply the conjugate expressions in such questions.

Using the difference of squares, \(\sin^4 x-\cos^4 x=(\sin^2 x-\cos^2 x)(\sin^2 x+\cos^2 x)\). Since \(\sin^2 x+\cos^2 x=1\), the simplified value is \(\sin^2 x-\cos^2 x\). Option B is its negative, so it is not correct. Exam tip: after applying \(a^2-b^2=(a-b)(a+b)\), use the fundamental trigonometric identity.

The numerator is (1+\tan^2 x=\sec^2 x) and the denominator is (1+\cot^2 x=\cosec^2 x). Their ratio is (\frac{\sec^2 x}{\cosec^2 x}=\tan^2 x).

Rationalising the denominator gives (\frac{(1+\sin x)^2}{\cos^2 x}). This equals ((\sec x+\tan x)^2).

By the half-angle identity, (\tan^2 \frac{x}{2}=\frac{1-\cos x}{1+\cos x}). Remember half-angle forms separately.

The standard half-angle identity is (\tan \frac{x}{2}=\frac{\sin x}{1+\cos x}). Identify the half-angle form quickly.

First, ((\sin x+\cos x)^2=1+2\sin x\cos x) gives (\sin x\cos x=\frac{5}{8}). Then ((\sin x-\cos x)^2=1-2\sin x\cos x=\frac{7}{4}).

Use \((\sin x-\cos x)^2=\sin^2x+\cos^2x-2\sin x\cos x=1-2\sin x\cos x\). Since \(\sin x-\cos x=\frac{1}{2}\), we get \(\frac{1}{4}=1-2\sin x\cos x\). Hence \(2\sin x\cos x=\frac{3}{4}\), so \(\sin x\cos x=\frac{3}{8}\). The value \(\frac{1}{4}\) is the square of the given difference, not the product. Exam tip: square the given expression first and then use \(\sin^2x+\cos^2x=1\).

Use the identity ((\sin x+\cos x)^2+(\sin x-\cos x)^2=2). Hence the value is (2-\frac{36}{25}=\frac{14}{25}).