Q.Prove that : $3\sin^{-1}x = \sin^{-1}(3x - 4x^3),\ x \in \left[-\dfrac{1}{2}, \dfrac{1}{2}\right]$

Showing the 12 most recent of 29 on this concept.

1. 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.

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

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2. 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 $4$. …

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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. …

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4. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.m, n and k are integers and $9.5 \leq n \leq 12$. If $\dfrac{(\cos\theta + i\sin\theta)^m}{(\sin\theta + i\cos\theta)^n} = k(\sin170^\circ - i\cos170^\circ)$, then $n - m - k =$ (A) $6$ (B) $12$ (C) $5$ (D) $7$

   ›Reveal solutionSolution

   The key is to rewrite both sides in the form $r(\cos\theta + i\sin\theta)$ using Euler’s formula and known trig identities, then equate moduli and arguments. The result is $n - m - k = 7$, so the correct option is (D).

   ---

   We start with the given equation:

   $\frac{(\cos\theta + i\sin\theta)^m}{(\sin\theta + i\cos\theta)^n} = k(\sin170^\circ - i\cos170^\circ)$

   where $m, n, k$ are integers and $9.5 \leq n \leq 12$.

   Concept & Intuition:

   The left side is a quotient of complex numbers in polar form. The right side is a complex number that can also be expressed in polar form. Our plan:

   • Convert everything to the form $r e^{i\phi}$ (or $r(\cos\phi + i\sin\phi)$).
   • Equate the modulus (gives $k$) and the argument (gives a relation between $m$, $n$, and $\theta$).
   • Use the integer constraint on $n$ to pin down the exact values.

   ---

   1. Rewrite the numerator and denominator in polar form.

   We know $(\cos\theta + i\sin\theta)^m = e^{i m\theta}$.

   For the denominator: $\sin\theta + i\cos\theta$.

   Notice that $\sin\theta = \cos(90^\circ - \theta)$ and $\cos\theta = \sin(90^\circ - \theta)$. So:

   $\sin\theta + i\cos\theta = \cos(90^\circ - \theta) + i\sin(90^\circ - \theta) = e^{i(90^\circ - \theta)}.$

   Thus the denominator is $[e^{i(90^\circ - \theta)}]^n = e^{i n(90^\circ - \theta)}$.

   Hence the left side becomes:

   $\frac{e^{i m\theta}}{e^{i n(90^\circ - \theta)}} = e^{i[m\theta - n(90^\circ - \theta)]} = e^{i[(m+n)\theta - 90^\circ n]}.$

   So the left side has modulus $1$ and argument $(m+n)\theta - 90^\circ n$.

   ---

   2. Rewrite the right side in polar form.

   We have $k(\sin170^\circ - i\cos170^\circ)$.

   First, note $\sin170^\circ = \sin(10^\circ)$ (since $\sin(180^\circ - x) = \sin x$) and $\cos170^\circ = -\cos10^\circ$. So:

   $\sin170^\circ - i\cos170^\circ = \sin10^\circ - i(-\cos10^\circ) = \sin10^\circ + i\cos10^\circ.$

   Now $\sin10^\circ + i\cos10^\circ = \cos(80^\circ) + i\sin(80^\circ)$ because $\sin10^\circ = \cos80^\circ$ and $\cos10^\circ = \sin80^\circ$. So:

   $\sin170^\circ - i\cos170^\circ = e^{i 80^\circ}.$

   Thus the right side is $k e^{i 80^\circ}$.

   ---

   3. Equate the two sides.

   We have:

   $e^{i[(m+n)\theta - 90^\circ n]} = k e^{i 80^\circ}.$

   Since the left side has modulus $1$, we must have $|k| = 1$. But $k$ is an integer, so $k = \pm 1$.

   Also, the arguments must be equal modulo $360^\circ$:

   $(m+n)\theta - 90^\circ n = 80^\circ + 360^\circ \cdot t, \quad t \in \mathbb{Z}.$

   ---

   4. Use the range of $n$ to determine $k$ and $\theta$.

   We are told $9.5 \leq n \leq 12$ and $n$ is an integer, so $n = 10, 11, 12$.

   We also need $\theta$ to be such that the equation holds for some integer $t$. Since $m$ and $n$ are integers, the left side’s argument depends on $\theta$. A natural choice is to pick $\theta$ so that the equation is simple. Often in such problems, $\theta$ is chosen to make $(m+n)\theta$ a multiple of $90^\circ$ or something clean.

   If we try $k=1$, then:

   $(m+n)\theta = 80^\circ + 90^\circ n + 360^\circ t.$

   For $n=10$: RHS = $80^\circ + 900^\circ = 980^\circ$, which mod $360^\circ$ is $260^\circ$. So $(m+10)\theta = 260^\circ + 360^\circ t$.

   For $n=11$: RHS = $80^\circ + 990^\circ = 1070^\circ$, mod $360^\circ$ is $350^\circ$.

   For $n=12$: RHS = $80^\circ + 1080^\circ = 1160^\circ$, mod $360^\circ$ is $80^\circ$.

   If $k=-1$, then $k e^{i80^\circ} = e^{i(80^\circ+180^\circ)} = e^{i260^\circ}$, so the argument becomes $260^\circ$ instead of $80^\circ$.

   ---

   5. Find a consistent integer solution.

   We also need $m$ to be an integer. Often the simplest case is when $\theta$ is a standard angle like $10^\circ$, $20^\circ$, etc.

   Try $\theta = 10^\circ$. Then $(m+n)\cdot10^\circ$ must equal one of the above.

   • For $n=10$, $k=1$: $(m+10)\cdot10 = 260 + 360t \Rightarrow m+10 = 26 + 36t$. For $t=0$, $m=16$ (integer). Works.
   • For $n=11$, $k=1$: $(m+11)\cdot10 = 350 + 360t \Rightarrow m+11 = 35 + 36t$. For $t=0$, $m=24$. …

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5. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0$ for $0 \leq \theta \leq \pi$, then $\theta =$ (A) $\dfrac{\pi}{3}, \dfrac{\pi}{6}, \dfrac{3\pi}{4}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}$ (B) $\dfrac{\pi}{2}, \dfrac{\pi}{6}, \dfrac{3\pi}{4}, \dfrac{7\pi}{6}$ (C) $\dfrac{\pi}{2}, \dfrac{\pi}{4}, \dfrac{\pi}{6}, \dfrac{3\pi}{4}, \dfrac{5\pi}{6}$ (D) $\dfrac{\pi}{2}, \dfrac{3\pi}{4}, \dfrac{5\pi}{6}, \dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The equation $\cos 6\theta + \cos 4\theta + \cos 2\theta + 1 = 0$ simplifies using sum-to-product identities and a double-angle substitution to a product of cosines equal to zero. The solutions in $[0,\pi]$ are $\theta = \frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{6}, \frac{3\pi}{4}, \frac{5\pi}{6}$, matching option (C).

   ---

   Concept & Intuition

   When you see a sum of cosines with angles in arithmetic progression (here $2\theta$, $4\theta$, $6\theta$), the natural reflex is to use the sum-to-product identity:

   $\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}.$

   Pairing terms cleverly collapses the sum into a product. Then the equation becomes a product of cosine factors equal to zero — each factor gives a family of angles. The only trick is to keep the domain $0 \le \theta \le \pi$ in mind and not accidentally include extraneous solutions.

   ---

   Step-by-step solution

   1. Pair the first two cosines

   $\cos 6\theta + \cos 4\theta = 2\cos\frac{6\theta+4\theta}{2}\cos\frac{6\theta-4\theta}{2} = 2\cos 5\theta \cos\theta.$

   So the equation becomes

   $2\cos 5\theta \cos\theta + \cos 2\theta + 1 = 0.$

   2. Rewrite $\cos 2\theta + 1$ using a double-angle identity Recall $\cos 2\theta = 2\cos^2\theta - 1$, so

   $\cos 2\theta + 1 = 2\cos^2\theta.$

   Now the equation is

   $2\cos 5\theta \cos\theta + 2\cos^2\theta = 0.$

   3. Factor out $2\cos\theta$

   $2\cos\theta \bigl( \cos 5\theta + \cos\theta \bigr) = 0.$

   So either $\cos\theta = 0$ or $\cos 5\theta + \cos\theta = 0$.

   4. Solve $\cos\theta = 0$ In $[0,\pi]$, $\cos\theta = 0$ gives

   $\theta = \frac{\pi}{2}.$

   5. Simplify $\cos 5\theta + \cos\theta = 0$ Use sum-to-product again:

   $\cos 5\theta + \cos\theta = 2\cos\frac{5\theta+\theta}{2}\cos\frac{5\theta-\theta}{2} = 2\cos 3\theta \cos 2\theta.$

   So $\cos 5\theta + \cos\theta = 0$ becomes

   $2\cos 3\theta \cos 2\theta = 0.$

   Hence either $\cos 3\theta = 0$ or $\cos 2\theta = 0$.

   6. Solve $\cos 2\theta = 0$ $\cos 2\theta = 0$ means $2\theta = \frac{\pi}{2} + k\pi$, i.e. $\theta = \frac{\pi}{4} + \frac{k\pi}{2}$. For $0 \le \theta \le \pi$, the values are

   $\theta = \frac{\pi}{4},\quad \frac{3\pi}{4}.$

   7. Solve $\cos 3\theta = 0$ …

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$\sin 9^\circ \sin 18^\circ \sin 36^\circ \sin 54^\circ \sin 72^\circ \sin 81^\circ =$ (A) $\dfrac{10 + 2\sqrt{5}}{128}$ (B) $\dfrac{5 - \sqrt{5}}{128}$ (C) $\dfrac{5 - \sqrt{5}}{64}$ (D) $\dfrac{10 - 2\sqrt{5}}{512}$

   ›Reveal solutionSolution

   Pair each sine with its complement to collapse the product to $\tfrac{1}{16}\sin^2 36^\circ=\dfrac{5-\sqrt5}{128}$ — option (B).

   We want

   $P=\sin 9^\circ\sin 18^\circ\sin 36^\circ\sin 54^\circ\sin 72^\circ\sin 81^\circ.$

   Use complements. $\sin 81^\circ=\cos 9^\circ,\ \sin 72^\circ=\cos 18^\circ,\ \sin 54^\circ=\cos 36^\circ$, so

   $P=(\sin 9^\circ\cos 9^\circ)(\sin 18^\circ\cos 18^\circ)(\sin 36^\circ\cos 36^\circ).$

   Double-angle on each pair ($\sin\theta\cos\theta=\tfrac12\sin 2\theta$):

   $P=\tfrac12\sin 18^\circ\cdot\tfrac12\sin 36^\circ\cdot\tfrac12\sin 72^\circ=\tfrac18\,\sin 18^\circ\sin 36^\circ\sin 72^\circ.$

   Collapse once more. Since $\sin 72^\circ=\cos 18^\circ$ and $\sin 18^\circ\cos 18^\circ=\tfrac12\sin 36^\circ$, …

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7. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In a triangle ABC, if $s = \frac{15}{2},\ a = 3,\ b = 5$ then $\frac{\sin \frac{B}{2}}{\sin \frac{A}{2}} =$ (A) $\cot \frac{C}{2}$ (B) $2 \sin \frac{C}{2}$ (C) $2 \csc \frac{C}{2}$ (D) $\sin \frac{C}{2}$

   ›Reveal solutionSolution

   With $c = 2s-a-b = 7$, the ratio $\dfrac{\sin(B/2)}{\sin(A/2)} = \sqrt{3} = 2\sin\dfrac{C}{2}$.

   Given $s=\dfrac{15}{2},\ a=3,\ b=5$, so the third side is

   $c = 2s - a - b = 15 - 3 - 5 = 7.$

   The half-angle sines are

   $\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}},\qquad \sin\frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{ac}}.$

   Taking the ratio, the common factor $(s-c)$ and $c$ cancel:

   $\frac{\sin(B/2)}{\sin(A/2)}=\sqrt{\frac{(s-a)\,b}{a\,(s-b)}}=\sqrt{\frac{\left(\tfrac{9}{2}\right)(5)}{3\left(\tfrac{5}{2}\right)}}=\sqrt{\frac{45/2}{15/2}}=\sqrt{3}.$

   Now express $\sqrt3$ through angle $C$. Since …

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8. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.In a triangle ABC, $2\sqrt{r_1 r_2 + r_2 r_3 + r_3 r_1} =$ (A) $8R \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2}$ (B) $8R \cos \dfrac{A}{2} \cos \dfrac{B}{2} \cos \dfrac{C}{2}$ (C) $4R \sin \dfrac{A}{2} \sin \dfrac{B}{2} \sin \dfrac{C}{2}$ (D) $4R \cos \dfrac{A}{2} \cos \dfrac{B}{2} \cos \dfrac{C}{2}$

   ›Reveal solutionSolution

   The expression $2\sqrt{r_1 r_2 + r_2 r_3 + r_3 r_1}$ simplifies to $8R \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}$, which matches option (B).

   The key here is to connect the exradii $r_1, r_2, r_3$ to the sides and the circumradius $R$ of triangle ABC. The expression under the square root — a symmetric sum of pairwise products of exradii — has a known compact form in terms of the semi-perimeter $s$, the area $\Delta$, and the sides. Once we express that sum, the square root collapses neatly into something involving $R$ and the half-angle cosines.

   Let’s go step by step.

   1. Recall the standard formulas for exradii. For a triangle with sides $a, b, c$, semi-perimeter $s$, and area $\Delta$, the exradii opposite vertices $A, B, C$ are:

   $r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c}.$

   These are always positive.

   2. Form the sum of pairwise products. We need $r_1 r_2 + r_2 r_3 + r_3 r_1$. Substituting:

   $r_1 r_2 = \frac{\Delta^2}{(s-a)(s-b)}, \quad r_2 r_3 = \frac{\Delta^2}{(s-b)(s-c)}, \quad r_3 r_1 = \frac{\Delta^2}{(s-c)(s-a)}.$

   So:

   $r_1 r_2 + r_2 r_3 + r_3 r_1 = \Delta^2 \left[ \frac{1}{(s-a)(s-b)} + \frac{1}{(s-b)(s-c)} + \frac{1}{(s-c)(s-a)} \right].$

   3. Combine the fractions. Put them over the common denominator $(s-a)(s-b)(s-c)$:

   $\frac{(s-c) + (s-a) + (s-b)}{(s-a)(s-b)(s-c)} = \frac{3s - (a+b+c)}{(s-a)(s-b)(s-c)}.$

   But $a+b+c = 2s$, so $3s - 2s = s$. Hence:

   $r_1 r_2 + r_2 r_3 + r_3 r_1 = \Delta^2 \cdot \frac{s}{(s-a)(s-b)(s-c)}.$

   4. Use the area formula to simplify. Heron’s formula gives $\Delta^2 = s(s-a)(s-b)(s-c)$. Therefore:

   $r_1 r_2 + r_2 r_3 + r_3 r_1 = \Delta^2 \cdot \frac{s}{(s-a)(s-b)(s-c)} = s(s-a)(s-b)(s-c) \cdot \frac{s}{(s-a)(s-b)(s-c)} = s^2.$

   That’s a beautifully clean result: the sum of pairwise products of exradii equals $s^2$.

   Tip

   This identity $r_1 r_2 + r_2 r_3 + r_3 r_1 = s^2$ is worth remembering — it saves a lot of time in problems involving exradii.

   5. Now take the square root. We have:

   $2\sqrt{r_1 r_2 + r_2 r_3 + r_3 r_1} = 2\sqrt{s^2} = 2s.$

   So the given expression is simply $2s$.

   6. Express $2s$ in terms of $R$ and the angles. The semi-perimeter $s$ can be written using the circumradius $R$ and the sines of the angles:

   $s = \frac{a+b+c}{2} = \frac{2R\sin A + 2R\sin B + 2R\sin C}{2} = R(\sin A + \sin B + \sin C).$

   There is a standard identity:

   $\sin A + \sin B + \sin C = 4 \cos\frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2}.$ …

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9. 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:

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

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10. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.If $\tan(60^\circ + \theta)\tan(60^\circ - \theta) = \dfrac{a \cos^2 \theta - b}{a \cos^2 \theta - c}$, then $\dfrac{a + c}{b} =$ (A) 3 (B) 5 (C) 7 (D) 9

    ›Reveal solutionSolution

    The key idea is to rewrite the product $\tan(60^\circ+\theta)\tan(60^\circ-\theta)$ using tangent addition formulas, simplify to an expression in $\cos^2\theta$, then match coefficients to find $\frac{a+c}{b}=3$.

    We start with the product

    $\tan(60^\circ+\theta)\tan(60^\circ-\theta).$

    A natural instinct is to use the formula $\tan(A+B)\tan(A-B) = \frac{\tan^2 A - \tan^2 B}{1 - \tan^2 A \tan^2 B}$, but here a more direct route is to write each tangent in terms of sine and cosine, then simplify. The goal is to express the product purely in terms of $\cos^2\theta$, so we can compare with the given form $\frac{a\cos^2\theta - b}{a\cos^2\theta - c}$.

    1. Write each tangent using sine and cosine

    $\tan(60^\circ+\theta) = \frac{\sin(60^\circ+\theta)}{\cos(60^\circ+\theta)},\quad \tan(60^\circ-\theta) = \frac{\sin(60^\circ-\theta)}{\cos(60^\circ-\theta)}.$

    Their product is

    $\frac{\sin(60^\circ+\theta)\sin(60^\circ-\theta)}{\cos(60^\circ+\theta)\cos(60^\circ-\theta)}.$

    2. Apply product-to-sum identities Recall:

    $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)],$

    $\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)].$

    Here $A=60^\circ+\theta$, $B=60^\circ-\theta$, so

    $A-B = 2\theta,\quad A+B = 120^\circ.$

    Thus

    $\sin(60^\circ+\theta)\sin(60^\circ-\theta) = \frac{1}{2}[\cos 2\theta - \cos 120^\circ],$

    $\cos(60^\circ+\theta)\cos(60^\circ-\theta) = \frac{1}{2}[\cos 2\theta + \cos 120^\circ].$

    3. Simplify using known values $\cos 120^\circ = -\frac{1}{2}$. So

    $\text{Numerator} = \frac{1}{2}\left[\cos 2\theta - \left(-\frac12\right)\right] = \frac{1}{2}\left(\cos 2\theta + \frac12\right),$

    $\text{Denominator} = \frac{1}{2}\left[\cos 2\theta + \left(-\frac12\right)\right] = \frac{1}{2}\left(\cos 2\theta - \frac12\right).$

    The factor $\frac12$ cancels in the ratio, giving

    $\tan(60^\circ+\theta)\tan(60^\circ-\theta) = \frac{\cos 2\theta + \frac12}{\cos 2\theta - \frac12}.$

    4. Convert $\cos 2\theta$ to $\cos^2\theta$ Use $\cos 2\theta = 2\cos^2\theta - 1$. Then $$$$

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11. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.The product of all the values of $\left(\sqrt{3} - i\right)^{\frac{3}{7}}$ is (A) $8$ (B) $-8$ (C) $8i$ (D) $-8i$

    ›Reveal solutionSolution

    The problem asks for the product of all distinct values of $(\sqrt{3} - i)^{3/7}$. The key is to interpret the exponent $3/7$ as a multivalued complex power, find all 7 distinct roots, and multiply them. The product simplifies to $-8i$, so the correct option is (D).

    ---

    Concept and Intuition

    When we raise a complex number to a fractional exponent like $3/7$, we are really solving for all complex numbers $z$ such that $z^7 = (\sqrt{3} - i)^3$. That is, we are taking the 7th roots of a fixed complex number. The product of all 7th roots of any nonzero complex number is always $0$? No — actually, the product of all $n$th roots of a complex number $w$ is $(-1)^{n-1} w$ (or equivalently, $(-1)^{n-1}$ times the original number). But here we have a twist: we are taking the 7th roots of $(\sqrt{3} - i)^3$, not of $(\sqrt{3} - i)$ itself. So we first compute that cube, then find all 7th roots, then multiply them.

    A classic pitfall: forgetting that the exponent $3/7$ means "cube first, then take 7th root" (or vice versa, but the set of values is the same). The product of all $n$th roots of a number $A$ is $(-1)^{n-1} A$. Here $n=7$, $A = (\sqrt{3} - i)^3$.

    ---

    Step-by-step solution

    1. Write $\sqrt{3} - i$ in polar form.

       Modulus: $|\sqrt{3} - i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2$.

       Argument: $\tan\theta = \frac{-1}{\sqrt{3}}$, and since the point $(\sqrt{3}, -1)$ is in quadrant IV, $\theta = -\frac{\pi}{6}$ (or $11\pi/6$).

       So $\sqrt{3} - i = 2 e^{-i\pi/6}$.

    2. Cube it.

       $(\sqrt{3} - i)^3 = (2 e^{-i\pi/6})^3 = 8 e^{-i\pi/2} = 8 (\cos(-\pi/2) + i\sin(-\pi/2)) = -8i$.

       So we need all 7th roots of $-8i$.

    3. Write $-8i$ in polar form.

       Modulus $8$, argument $-\pi/2$ (or $3\pi/2$). General argument: $-\pi/2 + 2k\pi$, $k \in \mathbb{Z}$.

       So $-8i = 8 e^{i(-\pi/2 + 2k\pi)}$.

    4. Find all 7th roots.

       The 7th roots are:

    $z_k = 8^{1/7} \, e^{i\left( \frac{-\pi/2 + 2k\pi}{7} \right)}, \quad k = 0,1,2,\dots,6.$

    Since $8^{1/7} = (2^3)^{1/7} = 2^{3/7}$, each root has modulus $2^{3/7}$.

    5. Product of all 7 roots.

       The product of all $n$th roots of a complex number $A$ is $(-1)^{n-1} A$. Here $n=7$, $A = -8i$.

       So product $= (-1)^{6} \cdot (-8i) = 1 \cdot (-8i) = -8i$.

       Alternatively, multiply the $z_k$ directly: …

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12. 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. …

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