Trigonometric functions and their properties

The direct answer is option C, 2π. A period is the smallest positive number T for which f(x+T)=f(x) wherever the function is defined. Since sec x=1/cos x and cos(x+2π)=cos x, we get sec(x+2π)=sec x. A smaller shift, π, changes neither cosine nor secant? Actually cos(x+π)=-cos x, so sec(x+π)=-sec x, not the same function. Thus 2π is the least positive period. Option A, π/π, equals 1 and is not the period. Option B, π/2, changes secant into a different value. Option C works exactly. Option D, 4π, is also a period, but it is not the fundamental or smallest period. Memory cue: secant follows cosine’s period, so remember 2π, not π.

Direct answer: the fundamental period of cot x is π, so option A is correct. A period T satisfies cot(x + T) = cot x wherever the function is defined. Since cot x = cos x/sin x, examine a shift of π. We know cos(x + π) = −cos x and sin(x + π) = −sin x. Therefore cot(x + π) = (−cos x)/(−sin x) = cos x/sin x = cot x. Thus π is a period. It is the least positive period because cotangent repeats after a half-turn, not only after a full turn. Option B, 2π, is also a period, but it is not the fundamental or smallest positive period. Option C, π/2, does not preserve cotangent values for general x. Option D, 3π, is another larger odd multiple of π, but again not the least period. Memory cue: tan and cot have fundamental period π, while sin and cos have period 2π. Values are undefined when sin x = 0, but that does not change the period.

Since \(\cosec x=\dfrac{1}{\sin x}\) and \(\sin(x+2\pi)=\sin x\), we get
\[\cosec(x+2\pi)=\cosec x\]
where defined. Also, the smallest positive shift that keeps \(\sin x\) unchanged (not just up to sign across undefined points) is \(2\pi\). Therefore, the period is \(2\pi\).

(\sec x=\frac{1}{\cos x}) and (\cos x) lies in ([-1,1]). Hence (\sec x) is greater than or equal to (1) or less than or equal to (-1).

\(\cosec x=\frac{1}{\sin x}\). Since \(-1\leq\sin x\leq 1\), and \(\sin x\neq0\) wherever cosecant is defined, its reciprocal satisfies \(\cosec x\leq-1\) or \(\cosec x\geq1\). Hence, the range is \((-\infty,-1]\cup[1,\infty)\). The interval \([-1,1]\) is the range of \(\sin x\), not of \(\cosec x\). Exam tip: reciprocal trigonometric functions cannot take values between \(-1\) and \(1\).

(\tan x=\frac{\sin x}{\cos x}), so it is undefined when (\cos x=0). A fraction is not defined when the denominator is zero.

(\cot x=\frac{\cos x}{\sin x}), so it is undefined when (\sin x=0). Always check the denominator carefully.

(\sec x=\frac{1}{\cos x}), so it is undefined when (\cos x=0). In reciprocal functions, the denominator cannot be zero.

(\cosec x=\frac{1}{\sin x}), so it is undefined when (\sin x=0). In a reciprocal, the denominator function must not be zero.

Using (\sin^2 x+\cos^2 x=1), (\cos x=\frac{4}{5}). In the first quadrant, (\cos x) is positive.

From (\sin^2 x=1-\cos^2 x), (\sin x=\frac{5}{13}). In the first quadrant, the positive value is taken.

Using (\sec^2 x=1+\tan^2 x), (\sec x=\frac{25}{24}). In the first quadrant, (\sec x) is positive.

From (\cosec^2 x=1+\cot^2 x), (\cosec x=\frac{17}{15}). Take the positive value in the first quadrant.

(\pi+x) lies in the third quadrant and (\sin x) is negative there. Hence (\sin(\pi+x)=-\sin x).

The direct answer is option B, −cos x. Use the angle-addition identity: cos(a+b)=cos a cos b−sin a sin b. Put a=π and b=x. Since cos π=−1 and sin π=0, cos(π+x)=(−1)cos x−0·sin x=−cos x. Option A, cos x, misses the sign change caused by adding π. Option B is exactly the identity result. Option C, sin x, is a different trigonometric function and has the wrong size and value in general. Option D, −sin x, also uses sine instead of cosine and is therefore wrong. Thinking in quadrants can help: adding π moves a point to the opposite side of the unit circle, so its x-coordinate, cosine, changes sign. The phrase “third quadrant” is not always enough unless x’s location is specified, but the identity is valid for every real x. Exam cue: adding π reverses both sine and cosine signs.

The correct identity is (1+\tan^2 x=\sec^2 x). In identity-based questions, check signs and fractions carefully.

The expression depends on tan^2 x, so first square the given value: tan^2 x = (1/2)^2 = 1/4. Substitute this into the numerator and denominator separately. The numerator becomes 1 - 1/4 = 3/4, while the denominator becomes 1 + 1/4 = 5/4. Hence the complete fraction is (3/4)/(5/4) = (3/4) × (4/5) = 3/5. Therefore option B is correct. The expression is also the standard identity cos 2x = (1 - tan^2 x)/(1 + tan^2 x), but finding x is unnecessary. Option A reverses the resulting ratio, while options C and D arise from incomplete or incorrect simplification. The denominator is nonzero for the supplied value, so the substitution is valid.

The governing definition is tan θ=sin θ/cos θ, provided cos θ is nonzero. The question directly gives tan θ=3, so the requested ratio sin θ/cos θ is immediately equal to 3. Therefore option B is correct. No triangle construction, quadrant analysis, or Pythagorean identity is needed because the required expression is exactly the definition of tangent. Option C reverses the ratio and represents cot θ, not tan θ. Option A incorrectly assumes that sine and cosine are equal, while option D would require sin θ=0, a condition not supplied and inconsistent with tan θ=3. The definition alone completely determines the answer.