Trigonometric functions and their properties

Given \(\tan x+\cot x=5\). Squaring both sides gives \((\tan x+\cot x)^2=\tan^2x+\cot^2x+2\tan x\cot x\). Since \(\tan x\cot x=1\), we get \(25=\tan^2x+\cot^2x+2\). Hence, \(\tan^2x+\cot^2x=23\). Option 25 is only the square of the given sum; the extra 2 must be subtracted. Exam tip: always use \(\tan x\cot x=1\) in such identities.

Using the identity \(\sec^2 x-\tan^2 x=1\), we get \((\sec x-\tan x)(\sec x+\tan x)=1\). Hence \(\frac{1}{4}(\sec x+\tan x)=1\), so \(\sec x+\tan x=4\). The option \(16\) may result from incorrectly squaring \(\frac{1}{4}\). Exam tip: rewrite \(\sec^2 x-\tan^2 x=1\) as a product in such questions.

The two expressions involving cosecant and cotangent are connected by a standard identity. Since \\(\cosec^2 x-\cot^2 x=1\\), their product can be written as \\(\bigl(\cosec x+\cot x\bigr)\bigl(\cosec x-\cot x\bigr)=1\\). This identity is valid wherever the expressions are defined, so it directly links the given sum with the required difference.

The given sum is \\(\cosec x+\cot x=6\\). Substituting this into the product identity gives \\(6(\cosec x-\cot x)=1\\). Dividing both sides by 6, we obtain \\(\cosec x-\cot x=\frac{1}{6}\\). Therefore, option A is correct. The answer is not 6 because the second factor must be the reciprocal of the first factor.

By the half-angle identity, (\frac{\sin x}{1-\cos x}=\cot \frac{x}{2}). In such forms, identify the denominator carefully.

The standard half-angle form is (\cot \frac{x}{2}=\frac{1+\cos x}{\sin x}). Keep the forms of (\tan \frac{x}{2}) and (\cot \frac{x}{2}) separate.

The range of \(\cos x\) is \([-1,1]\). Adding 2 to both endpoints gives the range \([1,3]\), so the minimum is 1. In exams, when a constant is added, shift both endpoints of the range by that constant.

Since \(\cos x\) lies between \(-1\) and \(1\), the minimum value of \(5\cos x\) is \(-5\). Hence, the minimum value of \(2+5\cos x\) is \(2-5=-3\), attained when \(\cos x=-1\). Option \(7\) is the maximum value, obtained when \(\cos x=1\). Exam tip: for \(a+b\cos x\) with \(b>0\), the minimum value is \(a-b\).

Since \(\sin 2x\) always lies between \(-1\) and \(1\), the range of \(3\sin 2x\) is \([-3,3]\). Adding \(-1\) shifts every value down by 1, giving \([-3-1,\,3-1]=[-4,2]\). Hence, option B is correct. Option A accounts only for multiplication by 3 and misses the downward shift by 1. Exam tip: the range of \(a\sin(bx)+c\) is \([c-|a|,\,c+|a|]\).

For a function of the form \(a\cos(bx)+c\), the period is \(\frac{2\pi}{|b|}\). Here, \(b=3\), so the period is \(\frac{2\pi}{3}\). The coefficient 2 changes only the amplitude, and +4 shifts the graph upward; neither changes the period. \(\frac{\pi}{3}\) is half of the period, so it is not correct. Exam tip: For \(\sin bx\) or \(\cos bx\), divide \(2\pi\) by \(|b|\) to find the period.

The governing property is that tangent has fundamental period π: tan(θ + π) = tan θ. For a transformed function tan(kx), the least positive period T must satisfy kT = π, so T = π/|k|. Here k = 2, giving T = π/2. Directly, tan(2(x + π/2)) = tan(2x + π) = tan(2x), which verifies that π/2 is a period. It is fundamental because no smaller positive shift can produce the required π change in the tangent argument for every x. Therefore option A is correct. π is also a period, but it is twice the least positive period; 2π is still larger, and 2π/3 does not produce a fixed π multiple in the argument. The multiplier 2 compresses the period rather than enlarging it.

In the second quadrant, (\sin x) is positive and (\cos x) is negative. Since (\sin x=\frac{4}{5}), (\tan x=-\frac{4}{3}).

In the third quadrant, (\cos x) is negative. Since (\cos x=-\frac{12}{13}), (\sec x=-\frac{13}{12}).

In the fourth quadrant, (\cos x) is positive and (\sin x) is negative. From the (7,24,25) triple, (\cos x=\frac{7}{25}).

(\cos x=-\frac{8}{17}). In the second quadrant, (\sin x) is positive, so (\sin x=\frac{15}{17}).

(2\pi-x) is related to the fourth quadrant. There (\sin x) is negative, so (\sin(2\pi-x)=-\sin x).

The direct answer is option C, cos x. Use the cosine subtraction identity: cos(A-B)=cos A cos B+sin A sin B. Take A=2π and B=x. Since cos 2π=1 and sin 2π=0, cos(2π-x)=1·cos x+0·sin x=cos x. Option A, -cos x, is wrong; that sign change occurs for cos(π-x), not this expression. Option B, sin x, is wrong because cosine does not become sine merely because an angle is subtracted. Option C works exactly. Option D, -sin x, is also wrong and confuses cosine with sine as well as introducing an unnecessary negative sign. Geometrically, subtracting x from one full revolution places the angle in the reflected position, whose cosine is the same. This identity is valid for every real x, so no quadrant restriction is needed. Memory cue: cosine is even around a full turn: cos(2π-x)=cos x.

The angles \(2\pi-x\) and \(-x\) are coterminal because they differ by \(2\pi\). Hence, \(\tan(2\pi-x)=\tan(-x)\). Since tangent is an odd function, \(\tan(-x)=-\tan x\). The \(\cot x\) option is incorrect because tangent changes to cotangent for complementary angles, not for this angle. Exam tip: First reduce \(2\pi-x\) to \(-x\).

In the form (\frac{3\pi}{2}+x), (\sin) changes to (\cos) with a negative sign. Hence the answer is (-\cos x).

(\frac{3\pi}{2}-x) is related to the third quadrant. (\cos) changes to (\sin) with a negative sign.

At (\frac{3\pi}{2}+x), (\tan) changes to (\cot) with a negative sign. Hence (\tan(\frac{3\pi}{2}+x)=-\cot x).

Using allied-angle identities, \(\sin(\pi-x)=\sin x\) and \(\cos(\pi+x)=-\cos x\). Therefore, \(\frac{\sin(\pi-x)}{\cos(\pi+x)}=\frac{\sin x}{-\cos x}=-\tan x\). \(\tan x\) is the closest distractor, but it misses the negative sign from the denominator. Exam tip: For angles involving \(\pi\pm x\), check the sign of each trigonometric function separately.

Using the identities \(\cos(2\pi-x)=\cos x\) and \(\sin(\pi+x)=-\sin x\),
\[\frac{\cos(2\pi-x)}{\sin(\pi+x)}=\frac{\cos x}{-\sin x}=-\cot x.\]
Therefore, the correct answer is \(-\cot x\). Choosing \(\cot x\) would ignore the negative sign in the denominator. Exam tip: for an angle of the form \(\pi+x\), sine has a negative sign.

The governing concept is the pair of Pythagorean identities sec^2 x = 1 + tan^2 x and cosec^2 x = 1 + cot^2 x. Substitute these two identities into the given sum: sec^2 x + cosec^2 x = (1 + tan^2 x) + (1 + cot^2 x). Combining like terms gives 2 + tan^2 x + cot^2 x. Therefore option B is correct. Option A is missing both constant terms, so it is smaller than the actual expression by 2. Option C uses first powers rather than the squares required by the identities and has no valid derivation here. Option D is also insufficient; tan x cot x may equal 1 where defined, but it cannot represent the complete sum. The identity-based substitution is valid wherever the original functions are defined.

(\tan x+\cot x=\frac{1}{\sin x\cos x}) and (\sec x\cosec x=\frac{1}{\sin x\cos x}). Hence the ratio is (1).

Put (\sec x=\frac{1}{\cos x}) and (\tan x=\frac{\sin x}{\cos x}). The ratio becomes (\frac{1}{\sin x}=\cosec x).