Q.Which of the following functions is decreasing on $\left(0, \dfrac{\pi}{2}\right)$?
(A) $\sin 2x$
(B) $\tan x$
(C) $\cos x$
(D) $\cos 3x$

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $\cos \theta = -\dfrac{3}{5}$ and $\pi < \theta < 3\pi/2$, then $\tan\left(\dfrac{\theta}{2}\right) =$ (A) $2$ (B) $-2$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   Use the half-angle formula for tangent in terms of cosine, and determine the sign of $\tan(\theta/2)$ from the quadrant of $\theta/2$. The value is $\boxed{-2}$.

   The key here is that $\tan(\theta/2)$ can be expressed directly from $\cos\theta$ using a standard identity, but the sign of the result depends on where $\theta/2$ lies. You cannot just plug numbers into a formula and take the positive root — the quadrant decides the sign.

   We are given $\cos\theta = -\frac{3}{5}$ and $\pi < \theta < \frac{3\pi}{2}$. That means $\theta$ is in the third quadrant, where both sine and cosine are negative. Now, what about $\theta/2$? Since $\theta$ is between $\pi$ and $1.5\pi$, dividing by 2 gives $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$. That puts $\theta/2$ in the second quadrant, where tangent is negative. So our final answer must be negative — that already eliminates options (A), (C), and (D), leaving only (B) as possible. But let’s verify.

   1. Recall the half-angle formula for tangent. There are several forms; the one that uses only $\cos\theta$ is:

   $\tan\frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$

   The sign is chosen based on the quadrant of $\theta/2$, not $\theta$.

   2. Substitute the given value.

   $\cos\theta = -\frac{3}{5}$

   So:

   $1 - \cos\theta = 1 - \left(-\frac{3}{5}\right) = 1 + \frac{3}{5} = \frac{8}{5}$

   $1 + \cos\theta = 1 + \left(-\frac{3}{5}\right) = 1 - \frac{3}{5} = \frac{2}{5}$

   Hence:

   $\frac{1 - \cos\theta}{1 + \cos\theta} = \frac{8/5}{2/5} = \frac{8}{2} = 4$

   So:

   $\tan\frac{\theta}{2} = \pm \sqrt{4} = \pm 2$

   3. Apply the sign from the quadrant. …

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2. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.If $y = \log_2 \sin x$, then the minimum value of $\cosh y$ is (A) $2$ (B) $\dfrac{2}{e}$ (C) $\dfrac{e}{2}$ (D) $1$

   ›Reveal solutionSolution

   The problem reduces to finding the minimum of $\cosh(\log_2 \sin x)$ over $x$ where $\sin x > 0$. Using the definition of $\cosh$ and properties of logs, this becomes a function of $\sin x$ alone; its minimum occurs when $\sin x = 1$, giving the answer $\boxed{1}$.

   The key here is to see that $\cosh y$ is defined as $\frac{e^y + e^{-y}}{2}$, and $y$ itself is $\log_2 \sin x$. So we are really composing a hyperbolic cosine with a logarithmic function of a trigonometric one. The domain is restricted: $\sin x$ must be positive for the log to be defined, so $x \in (0, \pi)$ plus periodic copies.

   Instead of jumping into calculus on a messy composite function, we can simplify algebraically first. That’s the smart move — let the structure of the expression do the work for you.

   1. Write $y = \log_2 \sin x$. Then $\cosh y = \frac{e^y + e^{-y}}{2}$.

   2. Substitute $y$:

      $e^y = e^{\log_2 \sin x} = 2^{\log_2 \sin x \cdot \log_2 e}$? That’s messy. Better: recall $a^{\log_b c} = c^{\log_b a}$. Here, $e^{\log_2 \sin x} = \sin x^{\log_2 e}$. But there’s an even cleaner path.

   3. Use the change-of-base: $\log_2 \sin x = \frac{\ln \sin x}{\ln 2}$. Then

      $e^y = e^{\frac{\ln \sin x}{\ln 2}} = (\sin x)^{1/\ln 2}$.

      Similarly, $e^{-y} = (\sin x)^{-1/\ln 2}$.

   4. So $\cosh y = \frac{1}{2} \left[ (\sin x)^{1/\ln 2} + (\sin x)^{-1/\ln 2} \right]$.

   Let $t = \sin x$. Since $x$ is real and we need $\sin x > 0$, we have $t \in (0, 1]$. The function becomes

   $f(t) = \frac{1}{2} \left( t^{a} + t^{-a} \right)$, where $a = \frac{1}{\ln 2} > 0$. …

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3. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The range of the real valued function $f(x) = \sin^{-1}\left(\sqrt{x^2 + x + 1}\right)$ is (A) $\left[ -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right]$ (B) $\left[ 0, \dfrac{\pi}{2} \right]$ (C) $\left[ \dfrac{\pi}{6}, \dfrac{\pi}{2} \right]$ (D) $\left[ \dfrac{\pi}{3}, \dfrac{\pi}{2} \right]$

   ›Reveal solutionSolution

   The key is that the argument of $\sin^{-1}$ must lie in $[-1,1]$, and the quadratic inside the square root has a minimum of $\frac34$, so the range of $\sqrt{x^2+x+1}$ is $[\frac{\sqrt3}{2}, \infty)$, but only $[\frac{\sqrt3}{2}, 1]$ is valid for $\sin^{-1}$. Hence the output range is $[\frac{\pi}{3}, \frac{\pi}{2}]$.

   Concept and intuition:

   The function $f(x) = \sin^{-1}\bigl(\sqrt{x^2 + x + 1}\bigr)$ is an inverse sine of a square root. Inverse sine only accepts inputs in $[-1, 1]$, and its output is in $[-\frac{\pi}{2}, \frac{\pi}{2}]$. But here the input is a square root, so it’s always non‑negative. That already restricts the output to $[0, \frac{\pi}{2}]$. The real question is: what are the actual possible values of $\sqrt{x^2 + x + 1}$ as $x$ runs over all real numbers? That will determine the exact subinterval of $[0, \frac{\pi}{2}]$ that $f$ can hit.

   1. Find the range of the quadratic inside the square root. The expression $x^2 + x + 1$ is a quadratic with positive leading coefficient. Its minimum occurs at $x = -\frac{1}{2}$:

   $\left(-\frac12\right)^2 + \left(-\frac12\right) + 1 = \frac14 - \frac12 + 1 = \frac34.$

   Since the quadratic opens upward, its range is $\left[\frac34, \infty\right)$.

   2. Apply the square root. Taking the square root preserves order for non‑negative numbers, so

   $\sqrt{x^2 + x + 1} \in \left[\sqrt{\frac34}, \infty\right) = \left[\frac{\sqrt3}{2}, \infty\right).$

   3. Intersect with the domain of $\sin^{-1}$. The inverse sine function $\sin^{-1}(t)$ is defined only for $t \in [-1, 1]$. Since our $t = \sqrt{x^2 + x + 1}$ is always $\ge \frac{\sqrt3}{2} \approx 0.866$, the only values that actually appear in the domain of $\sin^{-1}$ are those in the intersection: $\left[\frac{\sqrt3}{2}, \infty\right) \cap [-1, 1] = \left[\frac{\sqrt3}{2}, 1\right].$ …

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4. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $f(x) = \log(2x-3) - 2x^2 + 6x - 4$ is a real valued function then the interval in which $f$ is an increasing function is (A) $(-\infty, 2)$ (B) $\left(\dfrac{3}{2}, 2\right)$ (C) $(2, \infty)$ (D) $\left(\dfrac{3}{2}, \infty\right)$

   ›Reveal solutionSolution

   For a function to be increasing, its derivative must be positive. After finding the derivative and considering the domain, the function increases only on $\left(\frac{3}{2}, 2\right)$.

   The key idea is simple: a function is increasing where its derivative is positive. But we must also respect the domain — the logarithm $\log(2x-3)$ is only defined when its argument is positive, so $2x-3 > 0$, i.e., $x > \frac{3}{2}$. Any interval we consider must lie entirely within this domain.

   Now, to find where $f$ increases, we differentiate and solve $f'(x) > 0$.

   1. Differentiate $f(x)$ $f(x) = \log(2x-3) - 2x^2 + 6x - 4$ Using the chain rule: derivative of $\log(2x-3)$ is $\frac{2}{2x-3}$. So

   $f'(x) = \frac{2}{2x-3} - 4x + 6.$

   2. Set up the inequality for increasing We need $f'(x) > 0$:

   $\frac{2}{2x-3} - 4x + 6 > 0.$

   3. Combine into a single fraction Write $-4x + 6$ as $\frac{(-4x+6)(2x-3)}{2x-3}$:

   $\frac{2 + (-4x+6)(2x-3)}{2x-3} > 0.$

   Expand the numerator:

   $(-4x+6)(2x-3) = -8x^2 + 12x + 12x - 18 = -8x^2 + 24x - 18$.

   Add the $2$:

   $2 - 8x^2 + 24x - 18 = -8x^2 + 24x - 16.$

   Factor out $-8$:

   $-8(x^2 - 3x + 2) = -8(x-1)(x-2).$

   So the inequality becomes:

   $\frac{-8(x-1)(x-2)}{2x-3} > 0.$

   4. Simplify the sign analysis Since $-8$ is a negative constant, multiplying both sides by $-1$ (which flips the inequality) gives:

   $\frac{(x-1)(x-2)}{2x-3} < 0.$

   Now we only need to find where this rational expression is negative.

   5. Find critical points The numerator $(x-1)(x-2) = 0$ at $x=1$ and $x=2$. The denominator $2x-3 = 0$ at $x = \frac{3}{2}$. These three points divide the real line into intervals. But remember: the domain of $f$ is $x > \frac{3}{2}$, so we only care about intervals to the right of $\frac{3}{2}$. …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The period of the function $f(x) = \dfrac{2 \sin \left( \dfrac{\pi x}{3} \right) \cos \left( \dfrac{2\pi x}{5} \right)}{3 \tan \left( \dfrac{7\pi x}{2} \right) - 5 \sec \left( \dfrac{5\pi x}{3} \right)}$ is (A) 30 (B) 60 (C) 300 (D) 150

   ›Reveal solutionSolution

   The overall period is the LCM of the periods of the numerator and denominator, which gives $30$.

   Concept

   For a quotient of periodic functions, the fundamental period is the least common multiple (LCM) of the periods of the numerator and the denominator. For $\sin(kx)$ and $\cos(kx)$ the period is $\dfrac{2\pi}{|k|}$; for $\tan(kx)$ and $\sec(kx)$ it is $\dfrac{\pi}{|k|}$ and $\dfrac{2\pi}{|k|}$ respectively.

   Solution

   Component periods:

   • $\sin\!\left(\dfrac{\pi x}{3}\right)$: $\dfrac{2\pi}{\pi/3}=6$
   • $\cos\!\left(\dfrac{2\pi x}{5}\right)$: $\dfrac{2\pi}{2\pi/5}=5$
   • $\tan\!\left(\dfrac{7\pi x}{2}\right)$: $\dfrac{\pi}{7\pi/2}=\dfrac{2}{7}$ …

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6. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.The number of solutions of the equation $\cos 6x + \cos 4x + \cos 2x = -1$ in $[0, \pi]$ is (A) 4 (B) 3 (C) 6 (D) 5

   ›Reveal solutionSolution

   Substituting $t=\cos 2x$ reduces the equation to $2t(2t-1)(t+1)=0$, giving $\cos 2x=0,\tfrac12,-1$ and exactly $5$ solutions in $[0,\pi]$, option (D).

   Let $t=\cos 2x$. Using $\cos 4x=2t^2-1$ and $\cos 6x=4t^3-3t$:

   $(4t^3-3t)+(2t^2-1)+t=-1.$

   $4t^3+2t^2-2t-1=-1\;\Rightarrow\;4t^3+2t^2-2t=0\;\Rightarrow\;2t(2t^2+t-1)=0.$

   Factoring the quadratic, $2t(2t-1)(t+1)=0$, so

   $\cos 2x=0,\quad \cos 2x=\tfrac12,\quad \cos 2x=-1.$

   For $x\in[0,\pi]$ we have $2x\in[0,2\pi]$:

   • $\cos 2x=0$: $2x=\dfrac{\pi}{2},\dfrac{3\pi}{2}\Rightarrow x=\dfrac{\pi}{4},\dfrac{3\pi}{4}$ (2 solutions). …

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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If $a, b$ are real numbers and $\alpha$ is a real root of $x^2 + 12 + 3\sin(a + b x) + 6x = 0$ then the value of $\cos(a + b\alpha)$ for the least positive value of $a + b\alpha$ is (A) $-1$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\dfrac{1}{2}$ (D) $0$

   ›Reveal solutionSolution

   Completing the square forces $\sin(a+b\alpha)=-1$ and $\alpha=-3$; the least positive $a+b\alpha=\frac{3\pi}{2}$ gives $\cos(a+b\alpha)=0$. Option (D).

   Rewrite the equation as

   $(x+3)^2 + 3 + 3\sin(a+bx) = 0.$

   At a real root $x=\alpha$,

   $3\sin(a+b\alpha) = -\big[(\alpha+3)^2+3\big] \le -3,$

   so $\sin(a+b\alpha)\le -1$. Since $\sin\ge -1$, equality is forced:

   $\sin(a+b\alpha) = -1 \quad\text{and}\quad (\alpha+3)^2 = 0 \Rightarrow \alpha = -3.$ …

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $\tan A < 0$ and $\tan 2A = -\dfrac{4}{3}$, then $\cos 6A =$ (A) $\dfrac{117}{125}$ (B) $-\dfrac{117}{125}$ (C) $\dfrac{120}{169}$ (D) $-\dfrac{120}{169}$

   ›Reveal solutionSolution

   Use the double-angle identity for tangent to find $\tan A$, then use the sign condition to pick the correct quadrant; compute $\cos 2A$ and $\cos 6A$ via triple-angle or repeated double-angle formulas. The final value is $-\dfrac{117}{125}$, so the correct option is (B).

   We are given $\tan 2A = -\frac{4}{3}$ and $\tan A < 0$. The goal is $\cos 6A$. The key is to first find $\tan A$ (or directly $\cos 2A$) from the double-angle formula, then use the sign condition to resolve ambiguity, and finally express $\cos 6A$ in terms of $\cos 2A$.

   ---

   1. Relate $\tan 2A$ to $\tan A$

   The double-angle identity for tangent is:

   $\tan 2A = \frac{2\tan A}{1 - \tan^2 A}.$

   We are told $\tan 2A = -\frac{4}{3}$, so:

   $\frac{2\tan A}{1 - \tan^2 A} = -\frac{4}{3}.$

   2. Solve for $\tan A$

   Cross-multiply:

   $3(2\tan A) = -4(1 - \tan^2 A) \quad\Rightarrow\quad 6\tan A = -4 + 4\tan^2 A.$

   Bring all terms to one side:

   $4\tan^2 A - 6\tan A - 4 = 0.$

   Divide by 2:

   $2\tan^2 A - 3\tan A - 2 = 0.$

   This is a quadratic in $\tan A$. Factor:

   $(2\tan A + 1)(\tan A - 2) = 0.$

   So $\tan A = -\frac{1}{2}$ or $\tan A = 2$.

   3. Use the sign condition $\tan A < 0$

   Since $\tan A < 0$, we discard $\tan A = 2$ and keep:

   $\tan A = -\frac{1}{2}.$

   Watch out

   A common mistake is to forget the sign condition and pick the wrong root, leading to a different $\cos 2A$ and thus a wrong $\cos 6A$.

   4. Find $\cos 2A$ from $\tan A$

   We can use the identity:

   $\cos 2A = \frac{1 - \tan^2 A}{1 + \tan^2 A}.$

   With $\tan A = -\frac{1}{2}$, we have $\tan^2 A = \frac{1}{4}$. Then:

   $\cos 2A = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} = \frac{\frac{3}{4}}{\frac{5}{4}} = \frac{3}{5}.$ …

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9. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The number of solutions of the equation $\tan \theta + \cot 2\theta = 1$ lying in the interval $(-\pi, \pi)$ is (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   The key is to rewrite $\tan\theta + \cot 2\theta$ in terms of $\tan\theta$ using the double-angle identity for $\cot 2\theta$, simplify to a quadratic in $\tan\theta$, solve, and then count distinct solutions in $(-\pi, \pi)$ — there are 2 solutions.

   The equation mixes $\tan\theta$ and $\cot 2\theta$. The natural instinct is to express everything in terms of a single trigonometric function, and $\tan\theta$ is the most convenient choice because $\cot 2\theta$ has a neat double-angle formula in terms of $\tan\theta$.

   Recall that $\cot 2\theta = \frac{1}{\tan 2\theta}$, and $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$. So

   $\cot 2\theta = \frac{1 - \tan^2\theta}{2\tan\theta}.$

   This substitution will turn the equation into an algebraic one in $t = \tan\theta$, provided we are careful about where $\tan\theta$ is undefined (i.e., $\theta = \pm\frac{\pi}{2}$ in $(-\pi, \pi)$) and where $\cot 2\theta$ is undefined (i.e., $2\theta = n\pi$, or $\theta = \frac{n\pi}{2}$). We'll check those points separately after solving.

   1. Substitute and simplify. Let $t = \tan\theta$. Then the equation becomes

   $t + \frac{1 - t^2}{2t} = 1.$

   Multiply through by $2t$ (assuming $t \neq 0$ for now):

   $2t^2 + (1 - t^2) = 2t \quad \Rightarrow \quad t^2 + 1 = 2t.$

   So $t^2 - 2t + 1 = 0$, i.e., $(t - 1)^2 = 0$. Hence $t = 1$.

   2. Solve $\tan\theta = 1$ in $(-\pi, \pi)$.

      The general solution is $\theta = \frac{\pi}{4} + n\pi$, $n \in \mathbb{Z}$.

      In the interval $(-\pi, \pi)$, the values are:

      • For $n = 0$: $\theta = \frac{\pi}{4}$.
      • For $n = -1$: $\theta = \frac{\pi}{4} - \pi = -\frac{3\pi}{4}$.
      • For $n = 1$: $\theta = \frac{\pi}{4} + \pi = \frac{5\pi}{4}$, which is outside $(-\pi, \pi)$ since $\frac{5\pi}{4} > \pi$. So we have two candidate solutions: $\theta = \frac{\pi}{4}$ and $\theta = -\frac{3\pi}{4}$.

   3. Check for excluded points.

      We multiplied by $2t$, so we must check if $t = 0$ (i.e., $\theta = 0, \pm\pi$) could be a solution. Plug $\theta = 0$ into the original equation: $\tan 0 + \cot 0$ is undefined because $\cot 0$ blows up. Similarly, $\theta = \pm\pi$ gives $\tan(\pm\pi) = 0$ but $\cot(2\pi)$ or $\cot(-2\pi)$ is undefined. So no extra solutions there.

      Also check points where $\tan\theta$ is undefined: $\theta = \pm\frac{\pi}{2}$. At $\theta = \frac{\pi}{2}$, $\tan\theta$ is undefined, so it cannot satisfy. At $\theta = -\frac{\pi}{2}$, same issue. …

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10. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.If $a, b$ are real numbers and $\alpha$ is a real root of $x^2 + 12 + 3\sin(a + bx) + 6x = 0$ then the value of $\cos(a + b\alpha)$ for the least positive value of $a + b\alpha$ is (A) $0$ (B) $\dfrac{1}{2}$ (C) $-1$ (D) $\dfrac{1}{\sqrt{2}}$

    ›Reveal solutionSolution

    The equation is rewritten as a perfect square plus a sine term, forcing both to be zero for a real root. This gives $\alpha = -3$ and, for the least positive value of $a + b\alpha$, $\cos(a + b\alpha) = 0$.

    The given equation is $x^2 + 12 + 3\sin(a + bx) + 6x = 0$. At first glance, it mixes a quadratic in $x$ with a sine term that depends on $x$ through $a + bx$. That looks messy — but the trick is to notice that the quadratic part can be completed into a perfect square.

    Rewrite the quadratic terms: $x^2 + 6x + 12 = (x^2 + 6x + 9) + 3 = (x+3)^2 + 3$. So the equation becomes

    $(x+3)^2 + 3 + 3\sin(a + bx) = 0,$

    or

    $(x+3)^2 + 3\bigl[1 + \sin(a + bx)\bigr] = 0.$

    Now, $x$ is a real number, so $(x+3)^2 \ge 0$. Also, $\sin(\text{anything}) \ge -1$, so $1 + \sin(a + bx) \ge 0$. That means the left-hand side is a sum of two non-negative terms. For their sum to be zero, each term must individually be zero.

    1. From $(x+3)^2 = 0$, we get $x = -3$. Since $\alpha$ is a real root, $\alpha = -3$.

    2. From $1 + \sin(a + b\alpha) = 0$, we get $\sin(a + b\alpha) = -1$.

    So $a + b\alpha = a + b(-3) = a - 3b$ must satisfy $\sin(a - 3b) = -1$. …

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11. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.If $f(x) = x + \log\left(\frac{x-1}{x+1}\right)$ is a well-defined real valued function then $f$ is (A) monotonically decreasing function (B) monotonically increasing function (C) increasing in $(1, \infty)$ and decreasing in $(-\infty, -1)$ (D) decreasing in $(1, \infty)$ and increasing in $(-\infty, -1)$

    ›Reveal solutionSolution

    The function is defined only for $|x|>1$, and its derivative $f'(x)=\frac{x^2+1}{x^2-1}$ is always positive on both intervals, so $f$ is monotonically increasing on each piece. The correct option is (B).

    We need to decide the monotonicity of

    $f(x)=x+\log\left(\frac{x-1}{x+1}\right)$

    where it is a well-defined real-valued function. The logarithm requires its argument to be positive:

    $\frac{x-1}{x+1}>0.$

    Solving this inequality gives $x<-1$ or $x>1$. So the domain is $(-\infty,-1)\cup(1,\infty)$. The function is not defined between $-1$ and $1$, so we examine monotonicity separately on each interval.

    The natural tool is the derivative. If $f'(x)>0$ on an interval, $f$ is increasing there; if $f'(x)<0$, it is decreasing.

    1. Compute the derivative

    $f'(x)=1+\frac{d}{dx}\left[\log(x-1)-\log(x+1)\right].$

    Using $\frac{d}{dx}\log u = \frac{u'}{u}$,

    $f'(x)=1+\frac{1}{x-1}-\frac{1}{x+1}.$

    2. Simplify Combine the fractions:

    $\frac{1}{x-1}-\frac{1}{x+1}=\frac{(x+1)-(x-1)}{(x-1)(x+1)}=\frac{2}{x^2-1}.$

    Hence

    $f'(x)=1+\frac{2}{x^2-1}=\frac{x^2-1+2}{x^2-1}=\frac{x^2+1}{x^2-1}.$

    3. Sign of $f'(x)$ on the domain

       • Numerator: $x^2+1>0$ for all real $x$.
       • Denominator: $x^2-1>0$ when $|x|>1$, which is exactly our domain.

       Therefore, on both $(-\infty,-1)$ and $(1,\infty)$, we have $f'(x)>0$.

    4. Conclusion about monotonicity …

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12. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
    Q.If $540^\circ < \theta < 630^\circ$ and $\tan \theta = \dfrac{5}{12}$, then $\dfrac{\cos \dfrac{\theta}{2} - 5 \sin \dfrac{\theta}{2}}{\sqrt{-(12 \sec \theta + 5 \csc \theta)}} =$ (A) $-26$ (B) $26$ (C) $1$ (D) $-1$

    ›Reveal solutionSolution

    Fix the quadrant of $\theta$ (third) and of $\theta/2$ (fourth), evaluate numerator and denominator; both equal $\sqrt{26}$, so the expression is $1$.

    Locate $\theta$. Since $540^\circ < \theta < 630^\circ$, subtracting $360^\circ$ gives $180^\circ < \theta < 270^\circ$ — the third quadrant, where $\tan\theta = \dfrac{5}{12} > 0$ is consistent. Hence

    $\sin\theta = -\dfrac{5}{13}, \qquad \cos\theta = -\dfrac{12}{13}.$

    Denominator.

    $12\sec\theta + 5\csc\theta = 12\left(-\dfrac{13}{12}\right) + 5\left(-\dfrac{13}{5}\right) = -13 - 13 = -26,$

    so

    $\sqrt{-(12\sec\theta + 5\csc\theta)} = \sqrt{26}.$

    Locate $\theta/2$. From $540^\circ < \theta < 630^\circ$ we get $270^\circ < \dfrac{\theta}{2} < 315^\circ$ — the fourth quadrant, where $\cos\dfrac{\theta}{2} > 0$ and $\sin\dfrac{\theta}{2} < 0$. Using half-angle formulas with $\cos\theta = -\dfrac{12}{13}$:

    $\cos^2\dfrac{\theta}{2} = \dfrac{1+\cos\theta}{2} = \dfrac{1 - \tfrac{12}{13}}{2} = \dfrac{1}{26} \;\Rightarrow\; \cos\dfrac{\theta}{2} = \dfrac{1}{\sqrt{26}},$ …

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