Q.$3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, $x \in \left[\frac{1}{2}, 1\right]$ Write the following functions in the simplest form:

Showing the 12 most recent of 29 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.If $x = \sin 18^\circ$ and $y = \tan 22 \dfrac{1}{2}^\circ$, then $4x(4x+2) =$ (A) $(y+1)^2$ (B) $3y(y+1)$ (C) $y^2 + y$ (D) $y^2 + 2y + 3$

   ›Reveal solutionSolution

   The key idea is to evaluate $x = \sin 18^\circ$ and $y = \tan 22.5^\circ$ using known exact values, then simplify $4x(4x+2)$ to match one of the given expressions in $y$. The result is $y^2 + 2y + 3$, which corresponds to option (D).

   The problem asks for a relationship between $\sin 18^\circ$ and $\tan 22.5^\circ$ — two angles that appear in standard exact-value tables. The trick is to recall their exact forms, then do algebra cleanly.

   Why this approach works:

   Both $\sin 18^\circ$ and $\tan 22.5^\circ$ have neat exact values (involving $\sqrt{5}$ and $\sqrt{2}$ respectively). Once you substitute these, the expression $4x(4x+2)$ becomes a number. Then you evaluate each option in terms of $y$ and see which matches that number. Alternatively, you can express everything in terms of $y$ directly — but the numeric route is simpler and less error-prone.

   Let’s go step by step.

   1. Find $x = \sin 18^\circ$ exactly. A standard derivation (using the geometry of a regular pentagon or solving $\sin 5\theta = 0$) gives:

   $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$

   So $x = \frac{\sqrt{5} - 1}{4}$.

   2. Find $y = \tan 22.5^\circ$ exactly. Using the half-angle formula for tangent:

   $\tan 22.5^\circ = \tan\left(\frac{45^\circ}{2}\right) = \frac{1 - \cos 45^\circ}{\sin 45^\circ} = \frac{1 - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{2 - \sqrt{2}}{\sqrt{2}} = \sqrt{2} - 1$

   So $y = \sqrt{2} - 1$.

   3. Compute $4x(4x+2)$. First, $4x = 4 \cdot \frac{\sqrt{5} - 1}{4} = \sqrt{5} - 1$. Then $4x + 2 = (\sqrt{5} - 1) + 2 = \sqrt{5} + 1$. So:

   $4x(4x+2) = (\sqrt{5} - 1)(\sqrt{5} + 1) = (\sqrt{5})^2 - 1^2 = 5 - 1 = 4$

   The expression simplifies to $4$.

   4. Now evaluate each option in terms of $y = \sqrt{2} - 1$.

      Compute $y+1 = \sqrt{2}$, so $y+1$ is just $\sqrt{2}$.

      • Option (A): $(y+1)^2 = (\sqrt{2})^2 = 2$. Not $4$.
      • Option (B): $3y(y+1) = 3(\sqrt{2} - 1)(\sqrt{2}) = 3(2 - \sqrt{2}) = 6 - 3\sqrt{2} \approx 1.76$. Not $4$.
      • Option (C): $y^2 + y = (\sqrt{2} - 1)^2 + (\sqrt{2} - 1) = (2 - 2\sqrt{2} + 1) + \sqrt{2} - 1 = 3 - 2\sqrt{2} + \sqrt{2} - 1 = 2 - \sqrt{2} \approx 0.59$. Not $4$.
      • Option (D): $y^2 + 2y + 3 = (\sqrt{2} - 1)^2 + 2(\sqrt{2} - 1) + 3 = (2 - 2\sqrt{2} + 1) + 2\sqrt{2} - 2 + 3 = (3 - 2\sqrt{2}) + 2\sqrt{2} + 1 = 4$.

      Only option (D) gives $4$.

   Watch out

   A common mistake is to misremember $\sin 18^\circ$ as $\frac{\sqrt{5}+1}{4}$ (that’s $\cos 36^\circ$). Double-check: $\sin 18^\circ \approx 0.309$, and $\frac{\sqrt{5}-1}{4} \approx 0.309$, while $\frac{\sqrt{5}+1}{4} \approx 0.809$ — so the minus sign is crucial.

   Tip

   If you prefer to avoid numeric checking, you can also work backwards: notice $4x(4x+2) = 4$ means $x = \frac{\sqrt{5}-1}{4}$, and then try to express $4$ in terms of $y = \sqrt{2}-1$ by completing a square: $y^2 + 2y + 3 = (y+1)^2 + 2 = 2 + 2 = 4$. That’s a neat shortcut.

   ✓Final answer

   The correct option is (D).

2. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $6 \cos 2\theta + 2 \cos^2 \left( \dfrac{\theta}{2} \right) + 2 \sin^2 \theta = 0,\ -\pi < \theta < \pi$, then $\theta =$ (A) $\dfrac{\pi}{3}$ (B) $\dfrac{\pi}{3},\ \cos^{-1} \left( \dfrac{3}{5} \right)$ (C) $\cos^{-1} \left( \dfrac{3}{5} \right)$ (D) $\pm \dfrac{\pi}{3},\ \pm \left( \pi - \cos^{-1} \dfrac{3}{5} \right)$

   ›Reveal solutionSolution

   Use double-angle and half-angle identities to rewrite the equation entirely in terms of $\cos\theta$, then solve the resulting quadratic. The solutions in $(-\pi,\pi)$ are $\theta = \pm \frac{\pi}{3}$ and $\theta = \pm \bigl(\pi - \cos^{-1}\frac{3}{5}\bigr)$, matching option (D).

   The key is to express every trigonometric term using a single variable. The equation mixes $\cos 2\theta$, $\cos^2(\theta/2)$, and $\sin^2\theta$ — three different forms. But each can be written in terms of $\cos\theta$ using standard identities. That reduces the problem to a quadratic in $\cos\theta$, which we solve and then find all $\theta$ in the given interval.

   1. Rewrite each term using $\cos\theta$.

      • $\cos 2\theta = 2\cos^2\theta - 1$
      • $\cos^2\left(\frac{\theta}{2}\right) = \frac{1+\cos\theta}{2}$
      • $\sin^2\theta = 1 - \cos^2\theta$

      Substitute these into the equation:

   $6(2\cos^2\theta - 1) + 2\left(\frac{1+\cos\theta}{2}\right) + 2(1 - \cos^2\theta) = 0.$

   2. Simplify step by step.

      First term: $6(2\cos^2\theta - 1) = 12\cos^2\theta - 6$

      Second term: $2 \cdot \frac{1+\cos\theta}{2} = 1 + \cos\theta$

      Third term: $2(1 - \cos^2\theta) = 2 - 2\cos^2\theta$

      Adding them:

   $(12\cos^2\theta - 2\cos^2\theta) + \cos\theta + (-6 + 1 + 2) = 0$

   $10\cos^2\theta + \cos\theta - 3 = 0.$

   3. Solve the quadratic in $\cos\theta$.

   $10\cos^2\theta + \cos\theta - 3 = 0$

   Factor or use the quadratic formula:

   $\cos\theta = \frac{-1 \pm \sqrt{1 + 120}}{20} = \frac{-1 \pm 11}{20}.$

   So $\cos\theta = \frac{10}{20} = \frac{1}{2}$ or $\cos\theta = \frac{-12}{20} = -\frac{3}{5}$.

   4. Find all $\theta$ in $(-\pi, \pi)$.

      • For $\cos\theta = \frac{1}{2}$: the principal solutions are $\theta = \pm \frac{\pi}{3}$ (since $\cos$ is even and both lie in $(-\pi,\pi)$).
      • For $\cos\theta = -\frac{3}{5}$: let $\alpha = \cos^{-1}\left(-\frac{3}{5}\right)$. Since $\cos^{-1}$ of a negative gives an angle in $(\pi/2, \pi)$, we have $\alpha = \pi - \cos^{-1}\left(\frac{3}{5}\right)$. The other solution in $(-\pi,\pi)$ is $-\alpha = -\bigl(\pi - \cos^{-1}\frac{3}{5}\bigr)$.

      Thus the full solution set is $\theta = \pm \frac{\pi}{3},\ \pm \left(\pi - \cos^{-1}\frac{3}{5}\right)$.

   Watch out

   A common mistake is to forget the negative angle solutions. Since the interval is $(-\pi, \pi)$ and $\cos$ is even, every cosine value except $\pm 1$ gives two angles symmetric about $0$ — both must be included.

   ✓Final answer

   The correct option is (D): $\theta = \pm \frac{\pi}{3},\ \pm \left( \pi - \cos^{-1} \frac{3}{5} \right)$.

3. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Number of solutions of the equation $3^{2\sin^2 x} + 3^{2\cos^2 x} = 6$ lying in the interval $[-\pi, \pi]$ is (A) 2 (B) 4 (C) 3 (D) 1

   ›Reveal solutionSolution

   The key idea is to rewrite the equation using the identity $\sin^2 x + \cos^2 x = 1$, then substitute $t = 3^{2\sin^2 x}$ to get a quadratic. The solutions in $[-\pi, \pi]$ are $x = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}$, giving 4 solutions.

   The equation $3^{2\sin^2 x} + 3^{2\cos^2 x} = 6$ looks symmetric in $\sin^2 x$ and $\cos^2 x$. Since $\sin^2 x + \cos^2 x = 1$, the two exponents are linked: if one is $a$, the other is $2 - a$ (because $2\cos^2 x = 2(1 - \sin^2 x) = 2 - 2\sin^2 x$). This suggests a substitution that turns the exponential equation into an algebraic one.

   Let $t = 3^{2\sin^2 x}$. Then $2\cos^2 x = 2(1 - \sin^2 x) = 2 - 2\sin^2 x$, so $3^{2\cos^2 x} = 3^{2 - 2\sin^2 x} = \frac{3^2}{3^{2\sin^2 x}} = \frac{9}{t}$.

   The equation becomes:

   $t + \frac{9}{t} = 6$

   Multiply through by $t$ (note $t > 0$ always, since it's an exponential):

   $t^2 + 9 = 6t \quad \Rightarrow \quad t^2 - 6t + 9 = 0 \quad \Rightarrow \quad (t - 3)^2 = 0$

   So $t = 3$. That is, $3^{2\sin^2 x} = 3^1$, hence $2\sin^2 x = 1$, giving $\sin^2 x = \frac{1}{2}$.

   Now $\sin^2 x = \frac{1}{2}$ means $\sin x = \pm \frac{1}{\sqrt{2}}$. In the interval $[-\pi, \pi]$, we find all $x$ where sine takes these values.

   1. Solve $\sin x = \frac{1}{\sqrt{2}}$: The principal solutions are $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$. In $[-\pi, \pi]$, we also have the negative counterparts: $x = -\frac{3\pi}{4}$ (since $\sin(-\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$, wait — careful: $\sin(-\frac{3\pi}{4}) = -\frac{1}{\sqrt{2}}$, not $+\frac{1}{\sqrt{2}}$). Let's list systematically.

      For $\sin x = \frac{1}{\sqrt{2}}$ in $[-\pi, \pi]$:

      • $x = \frac{\pi}{4}$ (Quadrant I)
      • $x = \frac{3\pi}{4}$ (Quadrant II)
      • Also $x = -\frac{7\pi}{4}$ is outside the interval, so no.
      • What about $x = -\frac{5\pi}{4}$? That's less than $-\pi$, so no. So only two: $\frac{\pi}{4}$ and $\frac{3\pi}{4}$.

   2. Solve $\sin x = -\frac{1}{\sqrt{2}}$: In $[-\pi, \pi]$:

      • $x = -\frac{\pi}{4}$ (Quadrant IV)
      • $x = -\frac{3\pi}{4}$ (Quadrant III)
      • Also $x = \frac{5\pi}{4}$ is outside $[-\pi, \pi]$, so no. So two more: $-\frac{\pi}{4}$ and $-\frac{3\pi}{4}$.

   Thus we have four distinct solutions in $[-\pi, \pi]$: $x = \pm \frac{\pi}{4}, \pm \frac{3\pi}{4}$.

   Watch out

   A common mistake is to forget that $\sin^2 x = \frac{1}{2}$ gives both positive and negative sine values, and to only list the positive ones. Also, check that both $\frac{3\pi}{4}$ and $-\frac{3\pi}{4}$ lie inside $[-\pi, \pi]$ — they do, since $-\pi \leq -\frac{3\pi}{4} \leq \pi$.

   Tip

   Once you get $\sin^2 x = \frac{1}{2}$, you can directly write $\sin x = \pm \frac{1}{\sqrt{2}}$. The number of solutions in $[-\pi, \pi]$ for $\sin x = k$ (with $|k| < 1$) is always 2 if $k \neq 0$, so here two values of $k$ give $2 \times 2 = 4$ solutions. Quick check: $\sin x = \frac{1}{\sqrt{2}}$ gives 2 solutions, $\sin x = -\frac{1}{\sqrt{2}}$ gives 2 more.

   ✓Final answer

   The number of solutions is 4, which corresponds to option (B).

4. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.$1 + \cos x + \cos^2 x + \cos^3 x + \ldots$ to $\infty = 4 + 2\sqrt{3}$, then $x =$ (A) $\dfrac{n\pi}{6}$ (B) $(4n \pm 1)\dfrac{\pi}{3}$ (C) $(12n \pm 1)\dfrac{\pi}{6}$ (D) $(3n \pm 1)\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   The infinite geometric series sums to $\frac{1}{1-\cos x}$ only if $|\cos x|<1$, and setting that equal to $4+2\sqrt{3}$ gives $\cos x = \frac{\sqrt{3}}{2}$, so $x = 2n\pi \pm \frac{\pi}{6}$, which matches option (C) after rewriting.

   We are told that the infinite sum

   $1 + \cos x + \cos^2 x + \cos^3 x + \cdots = 4 + 2\sqrt{3}.$

   This is a geometric series with first term $1$ and common ratio $\cos x$. For the sum to converge to a finite number, we must have $|\cos x| < 1$ (otherwise the series diverges or is trivial). The sum to infinity is

   $S = \frac{1}{1 - \cos x}.$

   So the equation becomes

   $\frac{1}{1 - \cos x} = 4 + 2\sqrt{3}.$

   1. Solve for $\cos x$ Invert both sides:

   $1 - \cos x = \frac{1}{4 + 2\sqrt{3}}.$

   Rationalize the denominator:

   $\frac{1}{4 + 2\sqrt{3}} = \frac{1}{2(2 + \sqrt{3})} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{2(4 - 3)} = \frac{2 - \sqrt{3}}{2}.$

   So

   $1 - \cos x = \frac{2 - \sqrt{3}}{2}.$

   Hence

   $\cos x = 1 - \frac{2 - \sqrt{3}}{2} = \frac{2 - (2 - \sqrt{3})}{2} = \frac{\sqrt{3}}{2}.$

   2. Find the general solution for $x$ We know $\cos x = \frac{\sqrt{3}}{2}$ means

   $x = 2n\pi \pm \frac{\pi}{6}, \quad n \in \mathbb{Z}.$

   This is the standard form.

   3. Match with the given options
      • Option (A): $\dfrac{n\pi}{6}$ — too many values, includes angles where cosine is not $\sqrt{3}/2$.
      • Option (B): $(4n \pm 1)\dfrac{\pi}{3}$ — gives angles like $\pi/3$, $5\pi/3$, etc., whose cosine is $1/2$, not $\sqrt{3}/2$.
      • Option (C): $(12n \pm 1)\dfrac{\pi}{6}$ — let’s test: For $n=0$: $\pm \pi/6$ works. For $n=1$: $(12 \pm 1)\pi/6 = 13\pi/6$ or $11\pi/6$. $13\pi/6 = 2\pi + \pi/6$, cosine is $\sqrt{3}/2$; $11\pi/6 = 2\pi - \pi/6$, cosine also $\sqrt{3}/2$. So this matches $2n\pi \pm \pi/6$ exactly.
      • Option (D): $(3n \pm 1)\dfrac{\pi}{3}$ — gives angles like $\pi/3$, $2\pi/3$, etc., not matching.

   Tip

   The key insight: $2n\pi \pm \frac{\pi}{6}$ can be rewritten as $(12n \pm 1)\frac{\pi}{6}$ because $2n\pi = \frac{12n\pi}{6}$. This is a common trick in multiple-choice problems — they disguise the period.

   Watch out

   A common mistake is to forget the convergence condition $|\cos x| < 1$. Here $\cos x = \sqrt{3}/2 \approx 0.866$, so it’s fine. If the sum had been smaller than $1/2$, the series would diverge — always check.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.$\sqrt{3} \csc 20^\circ - \sec 20^\circ =$ (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   The expression $\sqrt{3} \csc 20^\circ - \sec 20^\circ$ simplifies to $4$ by converting to sines and cosines, using angle addition formulas, and recognizing a standard sine value.

   The key here is to rewrite everything in terms of sine and cosine, then look for a way to combine the terms into a single trigonometric ratio. The presence of $\sqrt{3}$ and $20^\circ$ hints at the sine addition formula — $\sin(60^\circ - \theta)$ or $\sin(60^\circ + \theta)$ often appears in such problems. The goal is to express the difference as a single fraction and simplify.

   1. Rewrite in terms of sine and cosine $\csc 20^\circ = \frac{1}{\sin 20^\circ}$ and $\sec 20^\circ = \frac{1}{\cos 20^\circ}$. So the expression becomes:

   $\sqrt{3} \cdot \frac{1}{\sin 20^\circ} - \frac{1}{\cos 20^\circ} = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}.$

   2. Combine into a single fraction The common denominator is $\sin 20^\circ \cos 20^\circ$:

   $\frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}.$

   3. Simplify the numerator using a known identity The numerator $\sqrt{3} \cos 20^\circ - \sin 20^\circ$ looks like $R \sin(60^\circ - 20^\circ)$ or $R \cos(60^\circ + 20^\circ)$. Recall:

   $\sin(60^\circ - \theta) = \sin 60^\circ \cos \theta - \cos 60^\circ \sin \theta = \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta.$

   Multiply both sides by 2:

   $2 \sin(60^\circ - \theta) = \sqrt{3} \cos \theta - \sin \theta.$

   Here $\theta = 20^\circ$, so:

   $\sqrt{3} \cos 20^\circ - \sin 20^\circ = 2 \sin(60^\circ - 20^\circ) = 2 \sin 40^\circ.$

   Tip

   A quick check: $\sin 40^\circ$ is positive, and the numerator is positive because $\sqrt{3} \cos 20^\circ \approx 1.732 \times 0.94 \approx 1.63$ and $\sin 20^\circ \approx 0.34$, so $2 \sin 40^\circ \approx 2 \times 0.643 \approx 1.286$ — matches.

   4. Simplify the denominator

      $\sin 20^\circ \cos 20^\circ = \frac{1}{2} \sin 40^\circ$ (using the double-angle identity $\sin 2\theta = 2 \sin \theta \cos \theta$). So the denominator becomes $\frac{1}{2} \sin 40^\circ$.

   5. Put it all together

      The expression is now:

   $\frac{2 \sin 40^\circ}{\frac{1}{2} \sin 40^\circ} = \frac{2 \sin 40^\circ}{1} \cdot \frac{2}{1 \cdot \sin 40^\circ} = 4.$

   The $\sin 40^\circ$ cancels (it is non-zero), leaving $4$.

   Watch out

   A common mistake is to forget the factor of 2 when converting $\sin 20^\circ \cos 20^\circ$ to $\frac{1}{2} \sin 40^\circ$, leading to an answer of 2 instead of 4. Always double-check the double-angle formula: $\sin 2\theta = 2 \sin \theta \cos \theta$, so $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$.

   ✓Final answer

   The value is $\boxed{4}$, which corresponds to option (D).

6. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.$4 \sin \frac{\pi}{6} \sin \frac{2\pi}{6} \sin \frac{3\pi}{6} \sin \frac{4\pi}{6} \sin \frac{5\pi}{6} =$ (A) $\cos \frac{\pi}{3} \cos \frac{2\pi}{3}$ (B) $\sin \frac{\pi}{3} \sin \frac{2\pi}{3}$ (C) $\sin \frac{\pi}{3} - \cos \frac{2\pi}{3}$ (D) $\cos \frac{\pi}{3} - \sin \frac{2\pi}{3}$

   ›Reveal solutionSolution

   The product equals $\tfrac{3}{4}$, which matches $\sin\tfrac{\pi}{3}\sin\tfrac{2\pi}{3} = \tfrac{3}{4}$.

   Evaluate each factor.

   $\sin\tfrac{\pi}{6}=\tfrac12,\quad \sin\tfrac{2\pi}{6}=\sin\tfrac{\pi}{3}=\tfrac{\sqrt3}{2},\quad \sin\tfrac{3\pi}{6}=\sin\tfrac{\pi}{2}=1,\quad \sin\tfrac{4\pi}{6}=\sin\tfrac{2\pi}{3}=\tfrac{\sqrt3}{2},\quad \sin\tfrac{5\pi}{6}=\tfrac12.$

   Multiply.

   $4 \cdot \tfrac12 \cdot \tfrac{\sqrt3}{2} \cdot 1 \cdot \tfrac{\sqrt3}{2} \cdot \tfrac12 = 4 \cdot \tfrac{1}{4} \cdot \tfrac{3}{4} = \tfrac{3}{4}.$

   Match the option. $\sin\tfrac{\pi}{3}\sin\tfrac{2\pi}{3} = \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3}{2} = \tfrac{3}{4}$, equal to our value. (Option A gives $-\tfrac14$, so it is rejected.)

   ✓Final answer

   The value is $\tfrac{3}{4} = \sin\tfrac{\pi}{3}\sin\tfrac{2\pi}{3}$ — option (B).

7. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.$2(\sin^6 \theta + \cos^6 \theta) - 3(\sin^4 \theta + \cos^4 \theta) =$ (A) $-1$ (B) $1$ (C) $0$ (D) $12$

   ›Reveal solutionSolution

   Reduce the sixth and fourth powers using $\sin^2\theta+\cos^2\theta=1$: $\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$ and $\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta$. The $\sin^2\theta\cos^2\theta$ terms cancel, leaving $-1$.

   Let $p=\sin^2\theta\cos^2\theta$ and use $\sin^2\theta+\cos^2\theta=1$.

   Fourth powers — with $a^2+b^2=(a+b)^2-2ab$ where $a=\sin^2\theta,\ b=\cos^2\theta$:

   $\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta=1-2p.$

   Sixth powers — with $a^3+b^3=(a+b)^3-3ab(a+b)$:

   $\sin^6\theta+\cos^6\theta=(\sin^2\theta+\cos^2\theta)^3-3\sin^2\theta\cos^2\theta(\sin^2\theta+\cos^2\theta)=1-3p.$

   Combine the two into the given expression:

   $2(1-3p)-3(1-2p)=2-6p-3+6p=(2-3)+(-6p+6p)=-1.$

   The $p$-terms cancel, so the expression is $-1$ for every $\theta$.

   ✓Final answer

   $2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)=-1$ — option (A).

8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.The equation that is satisfied by the general solution of the equation $4 - 3\cos^2\theta = 5\sin\theta\cos\theta$ is (A) $7\sin^2\theta + 3\cos^2\theta = 4$ (B) $\sin^2\theta - 2\cos\theta + \dfrac{1}{4} = 0$ (C) $\cot\theta - \tan\theta = \sec\theta$ (D) $1 + \sin^2\theta = 3\cos^2\theta$

   ›Reveal solutionSolution

   The equation factors as $(\sin\theta-\cos\theta)(4\sin\theta-\cos\theta)=0$. The $\sin\theta=\cos\theta$ family is exactly option (D) $1+\sin^2\theta=3\cos^2\theta$, the only option that reproduces a factor of the equation.

   Step 1 — clear the constant. Write $4=4(\sin^2\theta+\cos^2\theta)$:

   $4-3\cos^2\theta=5\sin\theta\cos\theta\ \Rightarrow\ 4\sin^2\theta+\cos^2\theta=5\sin\theta\cos\theta.$

   Step 2 — factor. Bringing all terms to one side,

   $4\sin^2\theta-5\sin\theta\cos\theta+\cos^2\theta=0\ \Rightarrow\ (\sin\theta-\cos\theta)(4\sin\theta-\cos\theta)=0,$

   so $\tan\theta=1$ or $\tan\theta=\tfrac14$.

   Step 3 — match the options. Dividing option (D) $1+\sin^2\theta=3\cos^2\theta$ by $\cos^2\theta$ gives $\sec^2\theta+\tan^2\theta=3\Rightarrow 2\tan^2\theta=2\Rightarrow\tan^2\theta=1$, i.e. $\sin\theta=\cos\theta$ — precisely the first factor of the general solution. The other options match no factor:

   • (A) $7\sin^2\theta+3\cos^2\theta=4\Rightarrow\sin^2\theta=\tfrac14$;
   • (B) carries a linear $\cos\theta$ term (non-homogeneous);
   • (C) reduces to $\cos2\theta=\sin\theta$.

   Hence the equation satisfied by the general solution is option (D).

   ✓Final answer

   $1+\sin^2\theta=3\cos^2\theta$ — option (D).

   ANSWER: D

9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $5\sin\theta + 3\cos\left(\theta + \dfrac{\pi}{3}\right) + 3$ lies between $\alpha$ and $\beta$ (including $\alpha$, $\beta$ also), then $(\alpha - \beta)(\alpha + \beta - 6) =$ (A) $28 - 5\sqrt{3}$ (B) $0$ (C) $3$ (D) $28 + 5\sqrt{3}$

   ›Reveal solutionSolution

   The expression simplifies to a single sine wave plus a constant; its range is $[3 - \sqrt{19}, 3 + \sqrt{19}]$, so $\alpha = 3 + \sqrt{19}$, $\beta = 3 - \sqrt{19}$, and $(\alpha - \beta)(\alpha + \beta - 6) = (2\sqrt{19})(6 - 6) = 0$.

   We start with the expression

   $E = 5\sin\theta + 3\cos\left(\theta + \frac{\pi}{3}\right) + 3.$

   The key idea: any linear combination of $\sin\theta$ and $\cos\theta$ can be written as $R\sin(\theta + \phi)$ or $R\cos(\theta + \phi)$, whose range is $[-R, R]$. Here we have a mix of $\sin\theta$ and a shifted cosine, so we first expand the cosine term to get everything in terms of $\sin\theta$ and $\cos\theta$, then combine them.

   ---

   1. Expand the cosine term

   $\cos\left(\theta + \frac{\pi}{3}\right) = \cos\theta\cos\frac{\pi}{3} - \sin\theta\sin\frac{\pi}{3} = \frac{1}{2}\cos\theta - \frac{\sqrt{3}}{2}\sin\theta.$

   So

   $3\cos\left(\theta + \frac{\pi}{3}\right) = \frac{3}{2}\cos\theta - \frac{3\sqrt{3}}{2}\sin\theta.$

   2. Combine with the $5\sin\theta$ term

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

   Let’s denote

   $A = 5 - \frac{3\sqrt{3}}{2}, \quad B = \frac{3}{2}.$

   So the non-constant part is $A\sin\theta + B\cos\theta$.

   3. Express as a single sine (or cosine) Any $A\sin\theta + B\cos\theta$ can be written as $R\sin(\theta + \phi)$ where

   $R = \sqrt{A^2 + B^2}.$

   Compute:

   $A^2 = \left(5 - \frac{3\sqrt{3}}{2}\right)^2 = 25 - 15\sqrt{3} + \frac{27}{4} = \frac{100}{4} - \frac{60\sqrt{3}}{4} + \frac{27}{4} = \frac{127 - 60\sqrt{3}}{4},$

   $B^2 = \frac{9}{4}.$

   Sum:

   $A^2 + B^2 = \frac{127 - 60\sqrt{3} + 9}{4} = \frac{136 - 60\sqrt{3}}{4} = 34 - 15\sqrt{3}.$

   So $R = \sqrt{34 - 15\sqrt{3}}$.

   Tip

   Simplify $\sqrt{34 - 15\sqrt{3}}$ by noticing it might be a perfect square of a binomial $a - b\sqrt{3}$.

   Suppose $(a - b\sqrt{3})^2 = a^2 + 3b^2 - 2ab\sqrt{3} = 34 - 15\sqrt{3}$.

   Then $a^2 + 3b^2 = 34$ and $2ab = 15$. Trying integers: $a=5$, $b=3/2$? No, $b$ must be rational. Try $a=5$, $b= \frac{5}{2}$? Then $2ab=25$, too big. Try $a= \frac{5}{2}$, $b=3$? Then $2ab=15$ works, and $a^2+3b^2 = \frac{25}{4} + 27 = \frac{25+108}{4} = \frac{133}{4} \neq 34$. Hmm.

   Actually, $34 - 15\sqrt{3} = \frac{1}{4}(136 - 60\sqrt{3})$, and we already saw $136 - 60\sqrt{3} = (10 - 3\sqrt{3})^2$? Check: $(10 - 3\sqrt{3})^2 = 100 + 27 - 60\sqrt{3} = 127 - 60\sqrt{3}$, not 136. So not that.

   Let’s solve directly: $a^2+3b^2=34$, $2ab=15$. From $b = \frac{15}{2a}$, substitute: $a^2 + 3\left(\frac{225}{4a^2}\right) = 34$. Multiply by $4a^2$: $4a^4 - 136a^2 + 675 = 0$. Let $u=a^2$: $4u^2 - 136u + 675 = 0$. Discriminant: $136^2 - 4\cdot4\cdot675 = 18496 - 10800 = 7696$, not a perfect square. So $R$ is not a nice surd — but we don’t need its exact form, only the range.

   4. Range of the expression Since $A\sin\theta + B\cos\theta$ ranges from $-R$ to $R$, the whole expression

   $E = (A\sin\theta + B\cos\theta) + 3$

   ranges from $3 - R$ to $3 + R$.

   So $\alpha = 3 + R$, $\beta = 3 - R$ (assuming $\alpha \ge \beta$).

   5. Compute the required product

   $\alpha - \beta = (3+R) - (3-R) = 2R,$

   $\alpha + \beta - 6 = (3+R + 3-R) - 6 = 6 - 6 = 0.$

   Therefore

   $(\alpha - \beta)(\alpha + \beta - 6) = (2R)(0) = 0.$

   Watch out

   A common mistake is to try to compute $R$ explicitly and then multiply, missing that $\alpha+\beta = 6$ exactly cancels the constant shift, making the second factor zero regardless of $R$.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.In a $\triangle ABC$, if $4a = b + c$, then $\tan \dfrac{B}{2} \tan \dfrac{C}{2} =$ (A) $\dfrac{1}{3}$ (B) $\dfrac{3}{5}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    The product $\tan\tfrac{B}{2}\tan\tfrac{C}{2}=\dfrac{s-a}{s}$. With $4a=b+c$ we get $s=\tfrac{5a}{2}$ and $s-a=\tfrac{3a}{2}$, so the product is $\tfrac35$.

    Using the half-angle formulas with semi-perimeter $s=\dfrac{a+b+c}{2}$:

    $\tan\frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}},\qquad \tan\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}.$

    Multiplying, the $(s-b)$ and $(s-c)$ factors cancel:

    $\tan\frac{B}{2}\tan\frac{C}{2}=\sqrt{\frac{(s-a)^2}{s^2}}=\frac{s-a}{s}.$

    Now apply $4a=b+c$, i.e. $b+c=4a$:

    $s=\frac{a+b+c}{2}=\frac{a+4a}{2}=\frac{5a}{2},\qquad s-a=\frac{b+c-a}{2}=\frac{4a-a}{2}=\frac{3a}{2}.$

    Hence

    $\tan\frac{B}{2}\tan\frac{C}{2}=\frac{s-a}{s}=\frac{3a/2}{5a/2}=\frac{3}{5}.$

    ✓Final answer

    $\tan\dfrac{B}{2}\tan\dfrac{C}{2}=\dfrac{3}{5}$ — option (B).

11. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.If $\alpha, \beta, 5$ are the roots of the equation $x^3 - a x + a = \dfrac{\sin^2 x + \cos^4 x}{\cos^2 x + \sin^4 x}$, then $a(\alpha + \beta) =$ (A) $-150$ (B) $-155$ (C) $75$ (D) $105$

    ›Reveal solutionSolution

    The key idea is that the right-hand side is actually constant (equal to 1), so the cubic reduces to $x^3 - a x + a = 1$, whose roots include 5; substituting $x=5$ gives $a=30$, then using sum of roots yields $\alpha+\beta = -5$, so $a(\alpha+\beta) = -150$.

    We start by noticing that the right-hand side looks like it might depend on $x$, but a clever simplification shows it is actually constant. That’s the crucial insight: once we realize the fraction equals 1 for all $x$, the equation becomes a pure polynomial equation, and we can use standard root relations.

    ---

    1. Simplify the trigonometric fraction.

    We have

    $\frac{\sin^2 x + \cos^4 x}{\cos^2 x + \sin^4 x}.$

    Use the identities $\sin^2 x = 1 - \cos^2 x$ and $\cos^2 x = 1 - \sin^2 x$ to rewrite numerator and denominator:

    • Numerator: $\sin^2 x + \cos^4 x = (1 - \cos^2 x) + \cos^4 x = 1 - \cos^2 x + \cos^4 x$.
    • Denominator: $\cos^2 x + \sin^4 x = (1 - \sin^2 x) + \sin^4 x = 1 - \sin^2 x + \sin^4 x$.

    Now notice that $1 - \cos^2 x + \cos^4 x = 1 - \cos^2 x(1 - \cos^2 x) = 1 - \cos^2 x \sin^2 x$.

    Similarly, $1 - \sin^2 x + \sin^4 x = 1 - \sin^2 x(1 - \sin^2 x) = 1 - \sin^2 x \cos^2 x$.

    Both numerator and denominator equal $1 - \sin^2 x \cos^2 x$, so the fraction is exactly $1$ for all $x$ (provided denominator is nonzero, which it always is since $\sin^2 x \cos^2 x \le \frac14$).

    Tip

    A quick check: at $x=0$, numerator = $0+1=1$, denominator = $1+0=1$. At $x=\pi/2$, same. So indeed constant.

    Thus the equation becomes

    $x^3 - a x + a = 1 \quad \Longrightarrow \quad x^3 - a x + (a-1) = 0.$

    ---

    2. Use the given root.

    We are told $\alpha, \beta, 5$ are the roots. So $x=5$ satisfies the cubic:

    $5^3 - a\cdot 5 + (a-1) = 0 \quad \Longrightarrow \quad 125 - 5a + a - 1 = 0.$

    Simplify: $124 - 4a = 0$, so $a = 31$.

    Watch out

    A common mistake is to forget the constant term shift: the original equation had $+a$ on the left, but after moving the 1 over, the constant becomes $a-1$, not $a$.

    ---

    3. Find $\alpha + \beta$ using sum of roots.

    For a cubic $x^3 + px^2 + qx + r = 0$, the sum of roots is $-p$. Here the cubic is

    $x^3 + 0\cdot x^2 - a x + (a-1) = 0,$

    so the sum of all three roots is $0$ (coefficient of $x^2$ is zero). Hence

    $\alpha + \beta + 5 = 0 \quad \Longrightarrow \quad \alpha + \beta = -5.$

    ---

    4. Compute $a(\alpha + \beta)$.

    We have $a = 31$ and $\alpha + \beta = -5$, so

    $a(\alpha + \beta) = 31 \times (-5) = -155.$

    ---

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.The circumference of the equilateral triangle having the three points $\theta_1, \theta_2, \theta_3$ lying on the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ as its vertices is $(r, s)$. Then the average of $\cos(\theta_1 - \theta_2)$, $\cos(\theta_2 - \theta_3)$ and $\cos(\theta_3 - \theta_1)$ is (A) $\dfrac{1}{2} \left[ \dfrac{3r^2}{a^2} + \dfrac{3s^2}{b^2} - 1 \right]$ (B) $\dfrac{3}{2} \left[ \dfrac{r^2}{a^2} + \dfrac{s^2}{b^2} \right]$ (C) $\dfrac{1}{3} \left[ \dfrac{r^2}{a^2} + \dfrac{s^2}{b^2} \right]$ (D) $\dfrac{1}{3} \left[ \dfrac{r^2}{a^2} + \dfrac{s^2}{b^2} + \dfrac{rs}{ab} \right]$

    ›Reveal solutionSolution

    The average of the cosines of the pairwise angle differences for an equilateral triangle inscribed in an ellipse is expressed in terms of the triangle’s circumradius components $r, s$. Using the parametric form of the ellipse and the condition for an equilateral triangle, the average simplifies to $\frac{1}{2}\left[\frac{3r^2}{a^2}+\frac{3s^2}{b^2}-1\right]$, which matches option (A).

    ---

    Concept and Intuition

    We have an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Any point on it can be written parametrically as $(a\cos\theta, b\sin\theta)$. The three vertices of the triangle are given by parameters $\theta_1, \theta_2, \theta_3$. The triangle is equilateral, so all sides equal. The problem gives the circumradius of this triangle as $(r, s)$ — but careful: this notation is ambiguous. In context, $r$ and $s$ likely refer to the circumradius in the $x$ and $y$ directions? Actually, more plausibly, the problem means the circumcircle of the equilateral triangle has center at $(r, s)$? No — re-reading: "The circumference of the equilateral triangle ... is $(r, s)$." That is odd phrasing. Most likely, it means the circumcenter of the triangle is at $(r, s)$. Yes: in many geometry problems, the circumcenter of an equilateral triangle inscribed in an ellipse is given coordinates $(r, s)$. So we treat $(r, s)$ as the circumcenter.

    We need the average of $\cos(\theta_1-\theta_2)$, $\cos(\theta_2-\theta_3)$, $\cos(\theta_3-\theta_1)$. For an equilateral triangle, the angular differences between vertices (in terms of the ellipse parameter) are not simply $120^\circ$ because the ellipse distorts angles. But we can use the geometry: the circumcenter is equidistant from all vertices, and the vertices satisfy the ellipse equation. This gives relations that let us express the cosines in terms of $r, s, a, b$.

    ---

    Step-by-step solution

    1. Parametric points and distances Let the vertices be

    $P_i = (a\cos\theta_i,\; b\sin\theta_i),\quad i=1,2,3.$

    The triangle is equilateral, so its circumcenter $(r, s)$ is equidistant from all three vertices. That means:

    $(a\cos\theta_i - r)^2 + (b\sin\theta_i - s)^2 = R^2 \quad \text{for } i=1,2,3,$

    where $R$ is the circumradius.

    2. Expand and rearrange Expanding:

    $a^2\cos^2\theta_i + b^2\sin^2\theta_i - 2a r\cos\theta_i - 2b s\sin\theta_i + r^2 + s^2 = R^2.$

    Using $\cos^2\theta_i = 1 - \sin^2\theta_i$ is not helpful; instead, note that $a^2\cos^2\theta_i + b^2\sin^2\theta_i$ is not constant. But we can rewrite using the ellipse equation: since each point lies on the ellipse, we have

    $\frac{(a\cos\theta_i)^2}{a^2} + \frac{(b\sin\theta_i)^2}{b^2} = \cos^2\theta_i + \sin^2\theta_i = 1,$

    which is automatically true. So that doesn't simplify directly.

    Instead, treat the three equations as:

    $a^2\cos^2\theta_i + b^2\sin^2\theta_i - 2a r\cos\theta_i - 2b s\sin\theta_i = R^2 - r^2 - s^2.$

    The left side varies with $i$, but the right side is constant. So the three expressions are equal.

    3. Use the equilateral property For an equilateral triangle, the circumcenter is also the centroid. So:

    $r = \frac{a(\cos\theta_1+\cos\theta_2+\cos\theta_3)}{3}, \quad s = \frac{b(\sin\theta_1+\sin\theta_2+\sin\theta_3)}{3}.$

    This is key: the centroid of the vertices equals the circumcenter.

    4. Express sums of cosines and sines From the above:

    $\cos\theta_1+\cos\theta_2+\cos\theta_3 = \frac{3r}{a},$

    $\sin\theta_1+\sin\theta_2+\sin\theta_3 = \frac{3s}{b}.$

    5. Square and add these sums Square both:

    $(\sum \cos\theta_i)^2 = \frac{9r^2}{a^2}, \quad (\sum \sin\theta_i)^2 = \frac{9s^2}{b^2}.$

    Add them:

    $\sum \cos^2\theta_i + \sum \sin^2\theta_i + 2\sum_{i<j}(\cos\theta_i\cos\theta_j + \sin\theta_i\sin\theta_j) = \frac{9r^2}{a^2} + \frac{9s^2}{b^2}.$

    But $\cos^2\theta_i + \sin^2\theta_i = 1$ for each $i$, so the first sum is $3$. Also,

    $\cos\theta_i\cos\theta_j + \sin\theta_i\sin\theta_j = \cos(\theta_i - \theta_j).$

    Hence:

    $3 + 2\left[ \cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1) \right] = \frac{9r^2}{a^2} + \frac{9s^2}{b^2}.$

    6. Solve for the average Let $A$ be the average of the three cosines:

    $A = \frac{\cos(\theta_1-\theta_2) + \cos(\theta_2-\theta_3) + \cos(\theta_3-\theta_1)}{3}.$

    From the equation above:

    $3 + 2 \cdot 3A = \frac{9r^2}{a^2} + \frac{9s^2}{b^2},$

    $3 + 6A = 9\left(\frac{r^2}{a^2} + \frac{s^2}{b^2}\right),$

    $6A = 9\left(\frac{r^2}{a^2} + \frac{s^2}{b^2}\right) - 3,$

    $A = \frac{3}{2}\left(\frac{r^2}{a^2} + \frac{s^2}{b^2}\right) - \frac{1}{2}.$

    Rewrite:

    $A = \frac{1}{2}\left[ \frac{3r^2}{a^2} + \frac{3s^2}{b^2} - 1 \right].$

    This matches option (A).

    Watch out

    A common mistake is to assume $\theta_1, \theta_2, \theta_3$ are equally spaced (like $0^\circ, 120^\circ, 240^\circ$) — that would be true only for a circle, not an ellipse. The parametric angles are not the geometric angles, so the centroid formula is essential.

    Tip

    The key trick: for any triangle, the centroid coordinates are the averages of the vertex coordinates. For an equilateral triangle, centroid = circumcenter. This gives the sums of cosines and sines directly.

    ✓Final answer

    The correct option is (A).

    ANSWER: A