Q.Show that the function given by $f(x) = \sin x$ is

Showing the 12 most recent of 16 on this concept.

1. 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 $\frac{3}{2}$.

   6. Test intervals within the domain

      • For $x$ between $\frac{3}{2}$ and $2$ (say $x=1.8$): $(x-1) > 0$, $(x-2) < 0$, so numerator is negative. Denominator $2x-3 > 0$. Negative divided by positive = negative. So the expression $< 0$ — condition satisfied.
      • For $x > 2$ (say $x=3$): $(x-1) > 0$, $(x-2) > 0$, numerator positive. Denominator positive. Positive divided by positive = positive. So $> 0$ — not increasing.

   Thus $f'(x) > 0$ only for $\frac{3}{2} < x < 2$.

   Watch out

   A common mistake is to forget the domain restriction $x > \frac{3}{2}$ and include $x<1$ or $1 < x < \frac{3}{2}$ where the function isn't even defined. Always check the domain first for logarithmic functions.

   ✓Final answer

   The function is increasing on $\left(\dfrac{3}{2}, 2\right)$, which corresponds to option (B).

2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\alpha = \tan \left( 2\sin^{-1} \left( \frac{2}{3} \right) \right)$ and $\beta = \sin \left( 2\tan^{-1} \left( \frac{1}{3} \right) \right)$, then the maximum value of $\alpha \sin \theta + \beta \cos \theta =$ (A) $1$ (B) $\dfrac{\sqrt{2009}}{5}$ (C) $\dfrac{\sqrt{2024}}{3}$ (D) $4\sqrt{5} + \dfrac{3}{5}$

   ›Reveal solutionSolution

   $\alpha=\tan(2\sin^{-1}\tfrac23)=4\sqrt5$ and $\beta=\sin(2\tan^{-1}\tfrac13)=\tfrac35$; the maximum is $\sqrt{\alpha^2+\beta^2}=\sqrt{2009}/5$.

   $\alpha$: with $\sin^{-1}\tfrac23=\phi$, $\tan\phi=\dfrac{2}{\sqrt5}$, so

   $\alpha=\tan2\phi=\frac{2\cdot\frac{2}{\sqrt5}}{1-\frac{4}{5}}=\frac{4/\sqrt5}{1/5}=4\sqrt5,\qquad \alpha^2=80.$

   $\beta$: with $\tan^{-1}\tfrac13=\psi$, $\tan\psi=\tfrac13$, so

   $\beta=\sin2\psi=\frac{2\cdot\frac13}{1+\frac19}=\frac{2/3}{10/9}=\frac35,\qquad \beta^2=\frac{9}{25}.$

   The maximum of $\alpha\sin\theta+\beta\cos\theta$ is $\sqrt{\alpha^2+\beta^2}$:

   $\sqrt{80+\tfrac{9}{25}}=\sqrt{\tfrac{2000+9}{25}}=\frac{\sqrt{2009}}{5}.$

   ✓Final answer

   $\dfrac{\sqrt{2009}}{5}$ — option (B).

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:

   $\left[\frac{\sqrt3}{2}, \infty\right) \cap [-1, 1] = \left[\frac{\sqrt3}{2}, 1\right].$

   4. Determine the output range of $f$. On $[\frac{\sqrt3}{2}, 1]$, $\sin^{-1}$ is increasing. So the smallest output is

   $\sin^{-1}\!\left(\frac{\sqrt3}{2}\right) = \frac{\pi}{3},$

   and the largest output is

   $\sin^{-1}(1) = \frac{\pi}{2}.$

   Hence the range of $f$ is $\left[\frac{\pi}{3}, \frac{\pi}{2}\right]$.

   Watch out

   A common mistake is to forget that the square root forces the argument of $\sin^{-1}$ to be non‑negative, so the output cannot be negative. That eliminates options like (A) and any interval starting below $0$.

   Tip

   The minimum of $x^2+x+1$ is $\frac34$, so the smallest possible input to $\sin^{-1}$ is $\frac{\sqrt3}{2}$. That immediately gives the left endpoint $\frac{\pi}{3}$ — no need to check other values.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $\sin A = -\dfrac{60}{61}$, $\cot B = -\dfrac{40}{9}$ and neither A nor B is in $4^{\text{th}}$ quadrant then $6 \cot A + 4 \sec B =$ (A) $\dfrac{26}{5}$ (B) $-\dfrac{26}{5}$ (C) $-3$ (D) $3$

   ›Reveal solutionSolution

   $A$ lies in the 3rd quadrant and $B$ in the 2nd, giving $\cot A = \tfrac{11}{60}$ and $\sec B = -\tfrac{41}{40}$, so $6\cot A + 4\sec B = -3$ — option (C).

   Locate the quadrants.

   • $\sin A = -\dfrac{60}{61} < 0 \Rightarrow A$ is in the 3rd or 4th quadrant; excluding the 4th, $A$ is in the 3rd quadrant.
   • $\cot B = -\dfrac{40}{9} < 0 \Rightarrow B$ is in the 2nd or 4th quadrant; excluding the 4th, $B$ is in the 2nd quadrant.

   Compute $\cot A$. In the 3rd quadrant cosine is negative:

   $\cos^2 A = 1 - \left(\frac{60}{61}\right)^2 = \frac{121}{3721}\Rightarrow \cos A = -\frac{11}{61},$

   $\cot A = \frac{\cos A}{\sin A} = \frac{-11/61}{-60/61} = \frac{11}{60}.$

   Compute $\sec B$. With $|\cot B| = \dfrac{40}{9}$, a reference triangle gives hypotenuse $\sqrt{40^2+9^2} = 41$. In the 2nd quadrant $\cos B < 0$, $\sin B > 0$:

   $\cos B = -\frac{40}{41},\qquad \sec B = \frac{1}{\cos B} = -\frac{41}{40}.$

   Evaluate the expression.

   $6\cot A + 4\sec B = 6\cdot\frac{11}{60} + 4\cdot\left(-\frac{41}{40}\right) = \frac{11}{10} - \frac{41}{10} = -\frac{30}{10} = -3.$

   ✓Final answer

   $6\cot A + 4\sec B = -3$ — option (C).

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:

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

   Numerator period $=\operatorname{LCM}(6,5)=30$.

   Denominator period $=\operatorname{LCM}\!\left(\dfrac{2}{7},\dfrac{6}{5}\right)=\dfrac{\operatorname{LCM}(2,6)}{\gcd(7,5)}=\dfrac{6}{1}=6$.

   Overall period $=\operatorname{LCM}(30,6)=30$.

   ✓Final answer

   $(A)\ 30$

6. 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

      Since the derivative is positive everywhere the function is defined, $f$ is strictly increasing on each interval of its domain. It is not defined on $(-1,1)$, so we cannot say it is increasing on the whole real line, but on each connected piece it is increasing. The statement “monotonically increasing function” in the context of a function with a disconnected domain usually means it is increasing on each interval of its domain — and that matches option (B).

   Watch out

   A common mistake is to forget the domain restriction and try to evaluate $f$ or $f'$ between $-1$ and $1$, where the function doesn’t exist. Also, note that “monotonically increasing” here applies separately to each branch; the function is not defined across the gap, so we don’t compare values from different branches.

   Tip

   The derivative simplifies beautifully to $\frac{x^2+1}{x^2-1}$, which is always positive for $|x|>1$ — no need to test points or second derivatives.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Consider the following statements Assertion (A): For $x \in \mathbb{R} - \{1\}$, $\dfrac{d}{dx} \left( \tan^{-1} \left( \dfrac{1+x}{1-x} \right) \right) = \dfrac{d}{dx} \left( \tan^{-1} x \right)$ Reason (R): For $x < 1$, $\tan^{-1} \left( \dfrac{1+x}{1-x} \right) = \dfrac{\pi}{4} + \tan^{-1} x$, for $x > 1$, $\tan^{-1} \left( \dfrac{1+x}{1-x} \right) = -\dfrac{3\pi}{4} + \tan^{-1} x$ The correct answer is (A) Both (A) and (R) are true, (R) is the correct explanation of (A) (B) Both (A) and (R) are true, (R) is not the correct explanation of (A) (C) (A) is true, but (R) is false (D) (A) is false, but (R) is true

   ›Reveal solutionSolution

   Both (A) and (R) are true, and (R) is the correct explanation of (A) — option (A).

   Since $\tan\!\left(\dfrac{\pi}{4}+\tan^{-1}x\right)=\dfrac{1+x}{1-x}$, taking principal values gives

   • for $x<1$: $\ \tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)=\dfrac{\pi}{4}+\tan^{-1}x$,
   • for $x>1$: $\ \tan^{-1}\!\left(\dfrac{1+x}{1-x}\right)=-\dfrac{3\pi}{4}+\tan^{-1}x$.

   So (R) is true. In each interval the expression differs from $\tan^{-1}x$ only by a constant, hence

   $\frac{d}{dx}\tan^{-1}\!\left(\frac{1+x}{1-x}\right)=\frac{d}{dx}\tan^{-1}x=\frac{1}{1+x^{2}}.$

   Thus (A) is true, and (R) is exactly why (A) holds.

   ✓Final answer

   (A) Both (A) and (R) are true, and (R) is the correct explanation of (A).

8. 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:

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

   From $\sin(a+b\alpha)=-1$, we get $a+b\alpha = -\frac{\pi}{2}+2n\pi$. The least positive value is

   $a+b\alpha = -\frac{\pi}{2}+2\pi = \frac{3\pi}{2},$

   so

   $\cos(a+b\alpha) = \cos\frac{3\pi}{2} = 0.$

   ✓Final answer

   $\cos(a+b\alpha) = 0$. Option (D).

9. 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 $\sin(a - 3b) = -1$.

   The general solution for $\sin\theta = -1$ is $\theta = -\frac{\pi}{2} + 2n\pi$, where $n$ is any integer. The solutions are $\ldots, -\frac{5\pi}{2}, -\frac{\pi}{2}, \frac{3\pi}{2}, \frac{7\pi}{2}, \ldots$ — so the least positive value of $a + b\alpha$ is $\frac{3\pi}{2}$ (obtained at $n=1$: $-\frac{\pi}{2} + 2\pi = \frac{3\pi}{2}$).

   The question asks for $\cos(a + b\alpha)$ at that least positive value. So compute:

   $\cos\left(\frac{3\pi}{2}\right) = 0.$

   Watch out

   A common mistake is to take $a + b\alpha = -\frac{\pi}{2}$ because it's the principal value, but that's negative. The problem explicitly asks for the least positive value, so you must take $n=1$ giving $\frac{3\pi}{2}$.

   Tip

   The key insight is recognising that the sum of two non-negative expressions equals zero forces each to be zero. This avoids any complicated trigonometric equation solving.

   ✓Final answer

   The value is $\boxed{0}$, which corresponds to option (A).

10. 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:

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

    Tip

    Alternatively, you could find $\sin 2A$ and $\cos 2A$ from $\tan 2A$ directly using a right triangle: $\tan 2A = -\frac{4}{3}$ means $\sin 2A = \pm \frac{4}{5}$ and $\cos 2A = \mp \frac{3}{5}$. But the sign of $\cos 2A$ depends on the quadrant of $2A$, which we can deduce from $\tan A < 0$ — that’s a bit more involved, so the $\tan A$ route is cleaner.

    5. Express $\cos 6A$ in terms of $\cos 2A$

    We use the triple-angle formula for cosine:

    $\cos 3\theta = 4\cos^3\theta - 3\cos\theta.$

    Let $\theta = 2A$, then $6A = 3(2A)$, so:

    $\cos 6A = 4\cos^3(2A) - 3\cos(2A).$

    6. Substitute $\cos 2A = \frac{3}{5}$

    $\cos 6A = 4\left(\frac{3}{5}\right)^3 - 3\left(\frac{3}{5}\right) = 4\left(\frac{27}{125}\right) - \frac{9}{5} = \frac{108}{125} - \frac{9}{5}.$

    Write $\frac{9}{5}$ with denominator 125: $\frac{9}{5} = \frac{225}{125}$. Then:

    $\cos 6A = \frac{108}{125} - \frac{225}{125} = -\frac{117}{125}.$

    Triple-angle cosine: $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$.

    7. Match with the options

    The value $-\frac{117}{125}$ corresponds to option (B).

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. 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 $\cos\theta = -\dfrac{12}{13}$:

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

    $\sin^2\dfrac{\theta}{2} = \dfrac{1-\cos\theta}{2} = \dfrac{1 + \tfrac{12}{13}}{2} = \dfrac{25}{26} \;\Rightarrow\; \sin\dfrac{\theta}{2} = -\dfrac{5}{\sqrt{26}}.$

    Numerator.

    $\cos\dfrac{\theta}{2} - 5\sin\dfrac{\theta}{2} = \dfrac{1}{\sqrt{26}} - 5\left(-\dfrac{5}{\sqrt{26}}\right) = \dfrac{1 + 25}{\sqrt{26}} = \dfrac{26}{\sqrt{26}} = \sqrt{26}.$

    Result.

    $\dfrac{\cos\tfrac{\theta}{2} - 5\sin\tfrac{\theta}{2}}{\sqrt{-(12\sec\theta + 5\csc\theta)}} = \dfrac{\sqrt{26}}{\sqrt{26}} = 1.$

    ✓Final answer

    The expression equals $1$ — option (C).

12. 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.

       Check where $\cot 2\theta$ is undefined: $2\theta = n\pi$ gives $\theta = \frac{n\pi}{2}$. In $(-\pi, \pi)$, these are $-\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi$. None of these are $\frac{\pi}{4}$ or $-\frac{3\pi}{4}$, so our candidates are safe.

    4. Verify the candidates.

       For $\theta = \frac{\pi}{4}$: $\tan\frac{\pi}{4} = 1$, $\cot\frac{\pi}{2} = 0$, so $1 + 0 = 1$. Works.

       For $\theta = -\frac{3\pi}{4}$: $\tan\left(-\frac{3\pi}{4}\right) = \tan\left(\pi - \frac{3\pi}{4}\right) = \tan\frac{\pi}{4} = 1$ (since tangent has period $\pi$), and $\cot\left(-\frac{3\pi}{2}\right) = \cot\left(-\frac{3\pi}{2}\right)$. Note $-\frac{3\pi}{2} = \frac{\pi}{2} - 2\pi$, so $\cot\left(-\frac{3\pi}{2}\right) = \cot\frac{\pi}{2} = 0$. So $1 + 0 = 1$. Works.

    Thus exactly two solutions exist in $(-\pi, \pi)$.

    Watch out

    A common mistake is to forget that $\tan\theta$ has period $\pi$, so $\tan(-\frac{3\pi}{4}) = \tan(\frac{\pi}{4}) = 1$, not $-1$. Also, always check domain restrictions — multiplying by $t$ assumes $t \neq 0$, and the original equation has undefined points that must be excluded.

    ✓Final answer

    The number of solutions is 2, so the correct option is (B).