Trigonometric functions and their properties

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 \(\sec^2 x\).

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}).

Combine the two fractions using the common denominator (1 + sin x)(1 - sin x). The numerator becomes (1 - sin x) + (1 + sin x) = 2. The denominator is a difference of squares: (1 + sin x)(1 - sin x) = 1 - sin^2 x = cos^2 x. Consequently, the expression simplifies to 2/cos^2 x = 2 sec^2 x, wherever the original denominators are defined. Thus option A is correct. Option C omits the factor 2 from the numerator. Option B incorrectly replaces cos^2 x with sin^2 x, confusing the relevant Pythagorean identity. Option D ignores the surviving denominator entirely. The decisive principle is 1 - sin^2 x = cos^2 x, followed by the reciprocal identity sec^2 x = 1/cos^2 x.

Use the common denominator (1 + cos x)(1 - cos x). Adding the numerators gives (1 - cos x) + (1 + cos x) = 2. The denominator simplifies by the difference-of-squares identity: (1 + cos x)(1 - cos x) = 1 - cos^2 x = sin^2 x. Therefore the whole expression is 2/sin^2 x = 2 cosec^2 x, wherever the original denominators are defined. Hence option C is correct. Option A uses the complementary identity in the wrong direction and gives a secant expression instead of a cosecant expression. Option B has the correct trigonometric type but misses the factor 2. Option D incorrectly removes the remaining denominator. The essential reasoning is algebraic combination followed by 1 - cos^2 x = sin^2 x.

Multiply the numerator and denominator by \(\sec x-\tan x\). The denominator becomes \((\sec x+\tan x)(\sec x-\tan x)=\sec^2x-\tan^2x=1\). Hence, \(\frac{1}{\sec x+\tan x}=\sec x-\tan x\). Option C is the negative of the required expression, so it is incorrect. Exam tip: use the identity \(\sec^2x-\tan^2x=1\) for reciprocal expressions of this form.

Direct answer: option C, \(\cosec x+\cot x\). Start with the meaning of reciprocal: \(1/(\cosec x-\cot x)\) is the number that multiplies \(\cosec x-\cot x\) to give 1. Using \(\cosec^2x-\cot^2x=1\), we factor the difference of squares: \((\cosec x-\cot x)(\cosec x+\cot x)=1\). Therefore the reciprocal is \(\cosec x+\cot x\), wherever the expressions are defined. Option A is \((\cosec x-\cot x)/(\cosec x-\cot x)=1\), so it is not generally the reciprocal. Option B similarly equals 1, because numerator and denominator are identical, so it is not the required expression. Option C gives \((\cosec x+\cot x)/1\), since the product with the original denominator is 1; hence it is correct. Option D is \((1/(\cosec x+\cot x))/(1/(\cosec x+\cot x))=1\), so it is also not the answer. Memory cue: to rationalise \(\cosec x-\cot x\), multiply by its conjugate \(\cosec x+\cot x\).

The direct answer is option C, 2. In a function of the form y=D+A sin(Bx), the amplitude is the absolute value |A|. It measures the maximum distance of the graph from its middle line y=D. Here y=6-2 sin 4x, so the middle line is y=6 and the sine coefficient is A=-2. Thus the amplitude is |-2|=2. Since -1≤sin 4x≤1, the function ranges from 6-2(1)=4 to 6-2(-1)=8; its midpoint is 6 and its distance to either extreme is 2. Option A, 6, is the vertical shift or midline, not amplitude. Option B, 4, is the lower extreme value, not amplitude. Option C works. Option D, 8, is the upper extreme value, not amplitude. The number 4 inside the sine affects period, not amplitude. Memory cue: amplitude is the absolute coefficient of sine or cosine.

The governing concept is the period of a transformed cosine function. The basic cosine function cos(kx) has period T = 2π/|k|. In 5 cos(x/2), the coefficient of x is k = 1/2, so T = 2π/(1/2) = 4π. The outside factor 5 changes only the amplitude: the graph is stretched vertically, but its horizontal repetition is unchanged. A direct check confirms the result: 5 cos((x + 4π)/2) = 5 cos(x/2 + 2π) = 5 cos(x/2). Thus 4π is a period, and it is the least positive one. Option C is correct. π and 2π would be appropriate only for different horizontal coefficients, while π/2 reverses the effect of the factor 1/2. Amplitude and period must not be confused.

Since \(\tan x=\frac{\sin x}{\cos x}\), it is undefined wherever \(\cos x=0\), i.e., at \(x=\frac{\pi}{2}+n\pi\). Its range is all real numbers, not \([-1,1]\). Exam tip: for a quotient trig function, first check where its denominator becomes zero.

Since \(\sin(-x)=-\sin x\), \(\sin x\) is an odd function, and its fundamental period is \(2\pi\). Although \(\tan x\) is also odd, its fundamental period is \(\pi\). In exams, verify each condition separately.

Since ((\sec x-\tan x)(\sec x+\tan x)=1), (\sec x+\tan x=\frac{5}{2}). Subtracting the two equations gives (\tan x=\frac{21}{20}).

The governing concept is the Pythagorean trigonometric identity sin^2 x + cos^2 x = 1. Factor the numerator: sin x − sin^3 x = sin x(1 − sin^2 x). Using 1 − sin^2 x = cos^2 x, the expression becomes [sin x · cos^2 x]/cos^2 x. Whenever cos x is not zero, cos^2 x cancels, leaving sin x. Therefore option C is correct. The restriction cos x ≠ 0 is necessary because the original expression would be undefined when cos x = 0. The alternatives cos x, tan x, and sec x do not follow from the factorisation and identity; they result from using an incorrect identity or cancelling incorrectly.

Since \(0\leq \cos^2 x\leq 1\), we get \(0\leq 4\cos^2 x\leq 4\). Hence, \(2-4\cos^2 x\) has minimum value \(-2\) when \(\cos^2 x=1\), and maximum value \(2\) when \(\cos^2 x=0\). Therefore, its range is \([-2,2]\). The interval \([0,2]\) is incorrect because the function can also take the value \(-2\). Exam tip: for an expression involving \(\cos^2 x\), first use its range \([0,1]\) and then test the endpoint values.

Because for complementary angles \( \sin\left(\frac{\pi}{2}-\theta\right)=\cos \theta\) and \( \cos\left(\frac{\pi}{2}-\theta\right)=\sin \theta\). In exams remember co-function formulas.

The fundamental trigonometric identity is \(\sin^2\theta+\cos^2\theta=1\). Therefore, \(k=1\). Option 2 may result from incorrectly treating both terms as individually equal to their maximum value; it is not true for every \(\theta\). Exam tip: whenever you see \(\sin^2\theta+\cos^2\theta\), replace it with 1.

Direct answer: option A, 1. The fundamental identity is \(\sec^2\theta=1+\tan^2\theta\). Rearranging gives \(\sec^2\theta-\tan^2\theta=1\). Therefore the expression simplifies to 1 wherever the trigonometric functions are defined. Option A is correct. Option B, \(\sin\theta\), is not generally 1; it changes with the angle. Option C, \(\cos\theta\), also changes with the angle and is not the fixed difference. Option D, 0, would incorrectly assume the two squares are equal, but the identity shows that \(\sec^2\theta\) is exactly one greater than \(\tan^2\theta\). One can also verify this from \(\sec\theta=1/\cos\theta\) and \(\tan\theta=\sin\theta/\cos\theta\), although the identity is quicker. Memory cue: remember the pair \(\sec^2\theta-\tan^2\theta=1\).