Trigonometric functions and their properties

(\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.

Use (\sec^2 x=1+\tan^2 x) and (\cosec^2 x=1+\cot^2 x). The sum becomes (2+\tan^2 x+\cot^2 x).

(\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).

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

Use the identity \(1+\cot^2 x=\cosec^2 x\). Thus, \(\sin^2 x(1+\cot^2 x)=\sin^2 x\cdot\cosec^2 x=\sin^2 x\cdot\frac{1}{\sin^2 x}=1\). Therefore, the correct answer is \(1\). \(\sin^2 x\) is only a factor, not the simplified value. Exam tip: When you see \(1+\cot^2 x\), recall the \(\cosec^2 x\) identity.

Using the identity \(1+\tan^2 x=\sec^2 x\), the expression becomes \(\cos^2 x\sec^2 x\). Since \(\sec^2 x=1/\cos^2 x\), its value is \(1\). \(\sec^2 x\) is only the value of the bracketed part, not of the complete expression. Exam tip: When you see \(1+\tan^2 x\), replace it with \(\sec^2 x\).

Put (\tan^2 x=\frac{1}{4}). Then the value is (\frac{1-\frac{1}{4}}{1+\frac{1}{4}}=\frac{3}{5}).

Substitute (\cot^2 x=\frac{9}{4}) and simplify. The value is (\frac{\frac{9}{4}-1}{\frac{9}{4}+1}=\frac{5}{13}).

Using \(a^2+b^2=(a+b)^2-2ab\), take \(a=\sin^2 x\) and \(b=\cos^2 x\). Then \(\sin^4 x+\cos^4 x=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x\). Since \(\sin^2 x+\cos^2 x=1\), the expression equals \(1-2\sin^2 x\cos^2 x\). Option A incorrectly has a plus sign instead of a minus sign. Exam tip: for fourth powers, treat \(\sin^2 x\) and \(\cos^2 x\) as the two terms before applying an algebraic identity.

Use (\sin^4 x+\cos^4 x=1-2\sin^2 x\cos^2 x). Since (\sin^2 x\cos^2 x=\frac{1}{16}), the value is (\frac{7}{8}).

Squaring gives (1+2\sin x\cos x=3), so (\sin x\cos x=1). Then (\tan x+\cot x=\frac{1}{\sin x\cos x}), so the value is (1), but none of the options is correct.

Squaring gives (1+2\sin x\cos x=\frac{49}{25}), so (\sin x\cos x=\frac{12}{25}). Now (\tan x+\cot x=\frac{1}{\sin x\cos x}=\frac{25}{12}).

Squaring gives (1-2\sin x\cos x=\frac{1}{9}). Thus (\sin x\cos x=\frac{4}{9}) and (\tan x+\cot x=\frac{9}{4}).

Since (\sec x-\tan x=\frac{1}{3}). Adding both equations gives (2\sec x=3+\frac{1}{3}), so (\sec x=\frac{5}{3}).

(\cosec x-\cot x=\frac{1}{4}). Subtracting gives (2\cot x=4-\frac{1}{4}), so (\cot x=\frac{15}{8}).

Combining denominators gives \(\frac{2}{1-\sin^2 x}\). This is \(\frac{2}{\cos^2 x}=2\sec^2 x\).