Q.Integrate the following function: $\sin (ax+b) \cos (ax+b)$

Showing the 12 most recent of 42 on this concept.

1. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $\sinh x = -\dfrac{4}{3}$ then $\sinh 2x + \cosh 2x =$ (A) $-\dfrac{31}{41}$ (B) $-\dfrac{20}{9}$ (C) $\dfrac{49}{41}$ (D) $\dfrac{1}{9}$

   ›Reveal solutionSolution

   Using the definitions of hyperbolic functions and the identity $\sinh 2x + \cosh 2x = e^{2x}$, we find $e^x$ from $\sinh x = -4/3$, then square to get $e^{2x}$, yielding the result $1/9$, which corresponds to option (D).

   We are given $\sinh x = -\dfrac{4}{3}$ and need $\sinh 2x + \cosh 2x$.

   The key insight: recall that $\sinh 2x + \cosh 2x = e^{2x}$ because

   $\sinh t = \frac{e^t - e^{-t}}{2}, \quad \cosh t = \frac{e^t + e^{-t}}{2}$

   so

   $\sinh t + \cosh t = e^t.$

   Thus the problem reduces to finding $e^{2x}$ from $\sinh x = -\dfrac{4}{3}$.

   1. Express $\sinh x$ in terms of $e^x$

   $\sinh x = \frac{e^x - e^{-x}}{2} = -\frac{4}{3}.$

   Multiply by 2:

   $e^x - e^{-x} = -\frac{8}{3}.$

   2. Let $u = e^x$ (so $u > 0$). Then $e^{-x} = 1/u$, and the equation becomes

   $u - \frac{1}{u} = -\frac{8}{3}.$

   Multiply through by $u$:

   $u^2 - 1 = -\frac{8}{3}u \quad\Rightarrow\quad u^2 + \frac{8}{3}u - 1 = 0.$

   3. Solve the quadratic Multiply by 3:

   $3u^2 + 8u - 3 = 0.$

   Discriminant: $\Delta = 8^2 - 4\cdot3\cdot(-3) = 64 + 36 = 100$.

   So

   $u = \frac{-8 \pm 10}{6}.$

   This gives $u = \frac{2}{6} = \frac{1}{3}$ or $u = \frac{-18}{6} = -3$.

   Since $u = e^x > 0$, we take $u = \frac{1}{3}$.

   4. Find $e^{2x}$ $$$$

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2. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The solution set of the equation $\cos^2 2x + \sin^2 3x = 1$ is (A) $\left\{ x / x = n\pi + \frac{\pi}{2}, n \in \mathbb{Z} \right\}$ (B) $\left\{ x / x = 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z} \right\}$ (C) $\left\{ x / x = \frac{n\pi}{5}, n \in \mathbb{Z} \right\}$ (D) $\left\{ x / x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z} \right\}$

   ›Reveal solutionSolution

   The equation $\cos^2 2x + \sin^2 3x = 1$ simplifies to $\sin^2 3x = \sin^2 2x$, which leads to two families of solutions; the union of these gives $x = \frac{n\pi}{5}$, so the correct option is (C).

   We start with the equation

   $\cos^2 2x + \sin^2 3x = 1.$

   A natural first thought is to use the identity $\cos^2 \theta = 1 - \sin^2 \theta$, but here the angles are different ($2x$ and $3x$). Instead, recall the Pythagorean identity: $\cos^2 \alpha + \sin^2 \alpha = 1$. Our equation looks similar, but the angles don’t match. That mismatch is the key: we can rewrite $\cos^2 2x$ as $1 - \sin^2 2x$, then the equation becomes

   $1 - \sin^2 2x + \sin^2 3x = 1 \quad\Rightarrow\quad \sin^2 3x - \sin^2 2x = 0.$

   So we have $\sin^2 3x = \sin^2 2x$. This is a clean, symmetric condition. Taking square roots gives $\sin 3x = \pm \sin 2x$, which is equivalent to two cases: $\sin 3x = \sin 2x$ or $\sin 3x = -\sin 2x$. But we can handle both elegantly using the identity $\sin A = \sin B$ or the difference-of-squares factorization.

   1. Rewrite using difference of squares

   $\sin^2 3x - \sin^2 2x = 0 \quad\Rightarrow\quad (\sin 3x - \sin 2x)(\sin 3x + \sin 2x) = 0.$

   So either $\sin 3x = \sin 2x$ or $\sin 3x = -\sin 2x$.

   2. Solve $\sin 3x = \sin 2x$ The general solution for $\sin A = \sin B$ is

   $A = B + 2n\pi \quad\text{or}\quad A = \pi - B + 2n\pi,\quad n\in\mathbb{Z}.$

   • First branch: $3x = 2x + 2n\pi \Rightarrow x = 2n\pi$.
   • Second branch: $3x = \pi - 2x + 2n\pi \Rightarrow 5x = \pi + 2n\pi \Rightarrow x = \frac{\pi}{5} + \frac{2n\pi}{5} = \frac{(2n+1)\pi}{5}$.

   3. Solve $\sin 3x = -\sin 2x$

      Note $-\sin 2x = \sin(-2x)$. So we have $\sin 3x = \sin(-2x)$. Again apply the same formula:

      • First branch: $3x = -2x + 2n\pi \Rightarrow 5x = 2n\pi \Rightarrow x = \frac{2n\pi}{5}$.
      • Second branch: $3x = \pi - (-2x) + 2n\pi = \pi + 2x + 2n\pi \Rightarrow x = \pi + 2n\pi$, i.e., $x = (2n+1)\pi$.

   4. Combine all solutions

      From step 2: $x = 2n\pi$ and $x = \frac{(2n+1)\pi}{5}$.

      From step 3: $x = \frac{2n\pi}{5}$ and $x = (2n+1)\pi$.

      Notice that $2n\pi$ and $(2n+1)\pi$ are just multiples of $\pi$, which are already included in $\frac{2n\pi}{5}$ when $n$ is a multiple of 5? Actually, let’s check:

      • $x = 2n\pi$ is $\frac{10n\pi}{5}$, which is of the form $\frac{2k\pi}{5}$ with $k=5n$. …

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3. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\tan A = \dfrac{2}{3}$, then $\sin 4A =$ (A) $\dfrac{8}{27}$ (B) $\dfrac{120}{169}$ (C) $\dfrac{144}{169}$ (D) $\dfrac{16}{27}$

   ›Reveal solutionSolution

   Use the double-angle identity for tangent to find $\tan 2A$, then use the sine double-angle formula in terms of tangent to get $\sin 4A$ directly. The result is $\frac{120}{169}$, which corresponds to option (B).

   We are given $\tan A = \frac{2}{3}$ and need $\sin 4A$. The direct approach: find $\sin 2A$ and $\cos 2A$ using tangent double-angle formulas, then use $\sin 4A = 2 \sin 2A \cos 2A$. Alternatively, we can compute $\tan 2A$ first and then express $\sin 4A$ in terms of $\tan 2A$ — that’s often cleaner because we avoid square roots.

   Concept & Intuition

   The key is that $\sin 4A$ can be written as $2 \sin 2A \cos 2A$, and both $\sin 2A$ and $\cos 2A$ can be expressed rationally in terms of $\tan A$ (or $\tan 2A$). Since $\tan A$ is given as a simple fraction, we can compute $\tan 2A$ exactly, then use the identity $\sin \theta = \frac{2 \tan(\theta/2)}{1+\tan^2(\theta/2)}$ with $\theta = 4A$ — but more directly, we use $\sin 4A = \frac{2 \tan 2A}{1+\tan^2 2A}$.

   Let’s go step by step.

   1. Find $\tan 2A$ using the double-angle formula

   $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \cdot \frac{2}{3}}{1 - \left(\frac{2}{3}\right)^2} = \frac{\frac{4}{3}}{1 - \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{5}{9}} = \frac{4}{3} \cdot \frac{9}{5} = \frac{36}{15} = \frac{12}{5}.$

   2. Express $\sin 4A$ in terms of $\tan 2A$ Recall the identity: for any angle $\theta$,

   $\sin \theta = \frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)}.$

   Here $\theta = 4A$, so $\theta/2 = 2A$. Thus

   $\sin 4A = \frac{2 \tan 2A}{1 + \tan^2 2A}.$

   3. Substitute $\tan 2A = \frac{12}{5}$ $\sin 4A = \frac{2 \cdot \frac{12}{5}}{1 + \left(\frac{12}{5}\right)^2} = \frac{\frac{24}{5}}{1 + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{25}{25} + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{169}{25}} = \frac{24}{5} \cdot \frac{25}{169} = \frac{24 \cdot 5}{169} = \frac{120}{169}.$ …

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4. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.

   [!FORMULA] $\frac{\sin 1^\circ + \sin 2^\circ + \ldots + \sin 89^\circ}{2(\cos 1^\circ + \cos 2^\circ + \ldots + \cos 44^\circ) + 1} =$

   (A) $2$ (B) $\frac{1}{\sqrt{2}}$ (C) $\frac{1}{2}$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   The key idea is to pair symmetric sine terms and use sum-to-product identities, leading to a telescoping simplification. The final value is $\frac{1}{\sqrt{2}}$, so the correct option is (B).

   ---

   Concept and intuition:

   We have a sum of sines from $1^\circ$ to $89^\circ$ in the numerator, and a sum of cosines from $1^\circ$ to $44^\circ$ (doubled, plus one) in the denominator. The symmetry $\sin(90^\circ - x) = \cos x$ suggests pairing terms. Also, the classic identity $\sin x + \sin(90^\circ - x) = \sqrt{2} \sin(x + 45^\circ)$ or, more directly, using sum-to-product, will collapse the numerator into something involving cosines. The denominator’s “$+1$” is a hint: $\cos 0^\circ = 1$, so we can think of it as $2\sum_{k=1}^{44} \cos k^\circ + \cos 0^\circ$, making a symmetric sum from $0^\circ$ to $44^\circ$ that pairs with the numerator’s structure.

   ---

   Step-by-step solution:

   1. Pair the sine terms symmetrically Notice $\sin 89^\circ = \cos 1^\circ$, $\sin 88^\circ = \cos 2^\circ$, …, $\sin 46^\circ = \cos 44^\circ$, and the middle term $\sin 45^\circ = \frac{1}{\sqrt{2}}$. So the numerator $S = \sin 1^\circ + \sin 2^\circ + \cdots + \sin 89^\circ$ becomes

   $S = (\sin 1^\circ + \sin 89^\circ) + (\sin 2^\circ + \sin 88^\circ) + \cdots + (\sin 44^\circ + \sin 46^\circ) + \sin 45^\circ.$

   2. Apply sum-to-product identity For any $x$, $\sin x + \sin(90^\circ - x) = 2 \sin 45^\circ \cos(45^\circ - x) = \sqrt{2} \cos(45^\circ - x)$. Thus each pair gives $\sqrt{2} \cos(45^\circ - k^\circ)$ for $k = 1, 2, \dots, 44$. So

   $S = \sqrt{2} \sum_{k=1}^{44} \cos(45^\circ - k^\circ) + \frac{1}{\sqrt{2}}.$

   3. Simplify the cosine sum As $k$ runs from $1$ to $44$, $(45^\circ - k^\circ)$ runs from $44^\circ$ down to $1^\circ$. So

   $\sum_{k=1}^{44} \cos(45^\circ - k^\circ) = \cos 44^\circ + \cos 43^\circ + \cdots + \cos 1^\circ = \sum_{k=1}^{44} \cos k^\circ.$

   Hence

   $S = \sqrt{2} \sum_{k=1}^{44} \cos k^\circ + \frac{1}{\sqrt{2}}.$

   4. Rewrite the denominator The denominator is $D = 2(\cos 1^\circ + \cdots + \cos 44^\circ) + 1$. Notice $1 = \cos 0^\circ$, so

   $D = 2\sum_{k=1}^{44} \cos k^\circ + \cos 0^\circ.$

   5. Form the ratio $$$$

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5. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.

   [!FORMULA] $\frac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} =$

   (A) $\cot \theta$ (B) $\cos 2\theta$ (C) $\tan \theta$ (D) $\tan 2\theta$

   ›Reveal solutionSolution

   Use double-angle identities to rewrite the numerator and denominator, then simplify to a single trigonometric ratio. The expression simplifies to $\tan \theta$, so the correct option is (C).

   The core idea here is that when you see $\cos 2\theta$ and $\sin 2\theta$ together, your first instinct should be to replace them with their basic forms in terms of $\sin \theta$ and $\cos \theta$. That turns a messy fraction into something algebraic you can factor and cancel.

   Let’s walk through it.

   1. Rewrite $\cos 2\theta$ and $\sin 2\theta$ using standard double-angle formulas.

      We have:

      $\cos 2\theta = 1 - 2\sin^2 \theta$ (or $\cos^2 \theta - \sin^2 \theta$, but the $1 - 2\sin^2 \theta$ form is handy here because of the $1$ in the numerator).

      Also, $\sin 2\theta = 2\sin \theta \cos \theta$.

   2. Substitute into the numerator.

      Numerator: $1 - \cos 2\theta + \sin 2\theta$

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

      $= 1 - 1 + 2\sin^2 \theta + 2\sin \theta \cos \theta$

      $= 2\sin^2 \theta + 2\sin \theta \cos \theta$

      $= 2\sin \theta (\sin \theta + \cos \theta)$.

   3. Substitute into the denominator.

      Denominator: $1 + \cos 2\theta + \sin 2\theta$

      Here, use $\cos 2\theta = 2\cos^2 \theta - 1$ (or $1 - 2\sin^2 \theta$, but the $2\cos^2 \theta - 1$ form pairs nicely with the $+1$).

      So: $1 + (2\cos^2 \theta - 1) + 2\sin \theta \cos \theta$

      $= 1 + 2\cos^2 \theta - 1 + 2\sin \theta \cos \theta$

      $= 2\cos^2 \theta + 2\sin \theta \cos \theta$

      $= 2\cos \theta (\cos \theta + \sin \theta)$.

   4. Form the fraction and simplify.

      The whole expression becomes:

      $\frac{2\sin \theta (\sin \theta + \cos \theta)}{2\cos \theta (\sin \theta + \cos \theta)}$ …

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6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $\sinh x = \dfrac{12}{5}$, then $\cosh 2x - \sinh 2x =$ (A) $25$ (B) $\dfrac{1}{125}$ (C) $\dfrac{1}{25}$ (D) $125$

   ›Reveal solutionSolution

   The key idea is to express $\cosh 2x - \sinh 2x$ in terms of $\sinh x$ using hyperbolic identities, then substitute $\sinh x = \frac{12}{5}$ to get $\frac{1}{25}$, which corresponds to option (C).

   We start with the given $\sinh x = \frac{12}{5}$. The expression we need is $\cosh 2x - \sinh 2x$. A direct approach using hyperbolic double-angle formulas will work cleanly.

   Concept and intuition:

   Recall the hyperbolic identities:

   • $\cosh 2x = \cosh^2 x + \sinh^2 x$
   • $\sinh 2x = 2 \sinh x \cosh x$

   The combination $\cosh 2x - \sinh 2x$ simplifies nicely because it resembles $( \cosh x - \sinh x )^2$. Indeed,

   $(\cosh x - \sinh x)^2 = \cosh^2 x - 2\sinh x \cosh x + \sinh^2 x = (\cosh^2 x + \sinh^2 x) - 2\sinh x \cosh x = \cosh 2x - \sinh 2x.$

   So the problem reduces to finding $\cosh x - \sinh x$, then squaring it.

   Step-by-step solution:

   1. Find $\cosh x$ from $\sinh x$. Use the fundamental identity: $\cosh^2 x - \sinh^2 x = 1$. Given $\sinh x = \frac{12}{5}$, we have

   $\cosh^2 x = 1 + \left(\frac{12}{5}\right)^2 = 1 + \frac{144}{25} = \frac{169}{25}.$

   Since $\cosh x \ge 1$ for real $x$, we take the positive root:

   $\cosh x = \sqrt{\frac{169}{25}} = \frac{13}{5}.$

   2. Compute $\cosh x - \sinh x$. $\cosh x - \sinh x = \frac{13}{5} - \frac{12}{5} = \frac{1}{5}.$ …

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7. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If A and B (A > B) are acute angles, $\sin(A-B) = \dfrac{16}{65}$ and $\sin B = \dfrac{5}{13}$ then $\tan A + \cot A =$ (A) $\dfrac{25}{12}$ (B) $\dfrac{12}{25}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{12}{5}$

   ›Reveal solutionSolution

   We use the given sine values to find $\cos A$ and $\sin A$ via angle subtraction and Pythagorean identities, then compute $\tan A + \cot A = \frac{25}{12}$. The correct option is (A).

   We are told $A$ and $B$ are acute angles with $A > B$, $\sin(A-B) = \frac{16}{65}$, and $\sin B = \frac{5}{13}$. We need $\tan A + \cot A$.

   Concept & Intuition

   The expression $\tan A + \cot A$ simplifies to $\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A} = \frac{\sin^2 A + \cos^2 A}{\sin A \cos A} = \frac{1}{\sin A \cos A}$. So if we can find $\sin A$ and $\cos A$, we are done. We know $\sin B$ and $\sin(A-B)$; we can find $\cos B$ and $\cos(A-B)$ using the Pythagorean identity (since angles are acute, all trig ratios are positive). Then using $\sin A = \sin[(A-B) + B]$, we can compute $\sin A$ and $\cos A$.

   Step-by-step solution

   1. Find $\cos B$

      Since $B$ is acute, $\cos B = \sqrt{1 - \sin^2 B} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$.

   2. Find $\cos(A-B)$

      $A-B$ is also acute (since $A>B$ and both acute, their difference is less than $90^\circ$). So

      $\cos(A-B) = \sqrt{1 - \sin^2(A-B)} = \sqrt{1 - \left(\frac{16}{65}\right)^2} = \sqrt{1 - \frac{256}{4225}} = \sqrt{\frac{3969}{4225}} = \frac{63}{65}$.

   3. Use the sine addition formula

      $\sin A = \sin[(A-B) + B] = \sin(A-B)\cos B + \cos(A-B)\sin B$

      Substitute:

      $\sin A = \frac{16}{65} \cdot \frac{12}{13} + \frac{63}{65} \cdot \frac{5}{13} = \frac{192}{845} + \frac{315}{845} = \frac{507}{845}$. …

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8. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.In any $\triangle ABC$, $\dfrac{b - c \cos A}{c - b \cos A} =$ (A) $\dfrac{\sin B}{\sin C}$ (B) $\dfrac{\cos C}{\cos B}$ (C) $\dfrac{\cos B}{\cos C}$ (D) $\dfrac{\sin C}{\sin B}$

   ›Reveal solutionSolution

   Use the projection formulas in a triangle: $b = c \cos A + a \cos C$ and $c = b \cos A + a \cos B$. Substituting these into the given expression simplifies directly to $\dfrac{\cos C}{\cos B}$.

   The key is to avoid jumping into sine rule expansions right away. In any triangle, the sides and angles are linked not only by the sine rule but also by projection formulas — these express each side as the sum of the projections of the other two sides onto it. They are elegant and often make problems like this collapse in one step.

   Let’s recall the projection formulas for a triangle $ABC$:

   • $a = b \cos C + c \cos B$
   • $b = c \cos A + a \cos C$
   • $c = a \cos B + b \cos A$

   These are derived by dropping altitudes and are perfectly valid for any triangle. They are also symmetric and easy to remember: each side equals the sum of the adjacent sides times the cosines of the included angles.

   Now, the expression we need to simplify is:

   $\frac{b - c \cos A}{c - b \cos A}$

   1. Rewrite the numerator using the projection formula for $b$:

   $b = c \cos A + a \cos C$

   So,

   $b - c \cos A = (c \cos A + a \cos C) - c \cos A = a \cos C$

   2. Rewrite the denominator using the projection formula for $c$:

   $c = a \cos B + b \cos A$

   So,

   $c - b \cos A = (a \cos B + b \cos A) - b \cos A = a \cos B$

   3. Form the ratio: $$$$

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9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.$\sin 20^\circ (4 + \sec 20^\circ) =$ (A) $\sqrt{3}$ (B) $-\sqrt{3}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   The key idea is to rewrite the expression using sine and cosine, then apply triple-angle identities to simplify it to a known constant. The final result is $\sqrt{3}$.

   We start with the expression

   $\sin 20^\circ (4 + \sec 20^\circ).$

   The presence of $\sec 20^\circ$ suggests rewriting everything in terms of sine and cosine, since $\sec \theta = 1/\cos \theta$. Then we look for a way to combine terms into a known trigonometric identity — the triple-angle formulas for sine and cosine are natural here because $20^\circ$ is one-third of $60^\circ$.

   ---

   1. Rewrite in terms of sine and cosine

   $\sin 20^\circ \left(4 + \frac{1}{\cos 20^\circ}\right) = 4\sin 20^\circ + \frac{\sin 20^\circ}{\cos 20^\circ} = 4\sin 20^\circ + \tan 20^\circ.$

   2. Express $\tan 20^\circ$ as $\frac{\sin 20^\circ}{\cos 20^\circ}$ and combine over a common denominator

   $4\sin 20^\circ + \frac{\sin 20^\circ}{\cos 20^\circ} = \frac{4\sin 20^\circ \cos 20^\circ + \sin 20^\circ}{\cos 20^\circ} = \frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ}.$

   3. Use the double-angle identity $2\sin 20^\circ \cos 20^\circ = \sin 40^\circ$, so $4\sin 20^\circ \cos 20^\circ = 2\sin 40^\circ$. Then the numerator becomes

   $2\sin 40^\circ + \sin 20^\circ.$

   4. Apply the triple-angle identity for sine Recall: $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. For $\theta = 20^\circ$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so

   $\frac{\sqrt{3}}{2} = 3\sin 20^\circ - 4\sin^3 20^\circ.$

   This doesn't directly match our numerator, but we can also use the identity for $\sin 3\theta$ in terms of products:

   $\sin 3\theta = 4\sin\theta \sin(60^\circ - \theta) \sin(60^\circ + \theta).$

   For $\theta = 20^\circ$, this gives

   $\sin 60^\circ = 4\sin 20^\circ \sin 40^\circ \sin 80^\circ.$

   So $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8}$.

   But we have $2\sin 40^\circ + \sin 20^\circ$ — not a product. So let's try a different path.

   5. Better approach: Use the triple-angle identity for cosine $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. For $\theta = 20^\circ$, $\cos 60^\circ = \frac{1}{2}$, so

   $\frac{1}{2} = 4\cos^3 20^\circ - 3\cos 20^\circ.$

   Multiply both sides by 2:

   $1 = 8\cos^3 20^\circ - 6\cos 20^\circ.$

   Rearranging:

   $8\cos^3 20^\circ - 6\cos 20^\circ - 1 = 0.$

   6. Relate this to our expression We have $\frac{2\sin 40^\circ + \sin 20^\circ}{\cos 20^\circ}$. Note $\sin 40^\circ = 2\sin 20^\circ \cos 20^\circ$, so

   $2\sin 40^\circ = 4\sin 20^\circ \cos 20^\circ.$

   Then numerator = $4\sin 20^\circ \cos 20^\circ + \sin 20^\circ = \sin 20^\circ (4\cos 20^\circ + 1)$.

   So the whole expression is

   $\frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ} = \tan 20^\circ (4\cos 20^\circ + 1).$

   7. Now use the triple-angle cosine identity From $8\cos^3 20^\circ - 6\cos 20^\circ = 1$, divide by $\cos 20^\circ$ (nonzero):

   $8\cos^2 20^\circ - 6 = \frac{1}{\cos 20^\circ}.$

   So $\frac{1}{\cos 20^\circ} = 8\cos^2 20^\circ - 6$.

   But we have $4\cos 20^\circ + 1$, not $1/\cos 20^\circ$. Let's instead multiply the triple-angle identity by something clever.

   8. A cleaner trick: Multiply numerator and denominator by something Consider the original expression:

   $\sin 20^\circ (4 + \sec 20^\circ) = \frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ}.$

   Now use the identity $\sin 20^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{2\cos 20^\circ + 1}$? Not directly.

   Instead, recall the triple-angle formula for sine in product form:

   $\sin 60^\circ = 4\sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = 4\sin 20^\circ \sin 40^\circ \sin 80^\circ.$

   So $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8}$. …

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10. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.In a triangle ABC, if $a = 5$, $b = 3$, $c = 7$, then $\sqrt{\dfrac{\sin(A-B)}{\sin(A+B)}} =$ (A) $\dfrac{4}{7}$ (B) $16$ (C) $36$ (D) $\dfrac{4}{5}$

    ›Reveal solutionSolution

    Using the law of sines and the tangent half‑angle formula, the expression simplifies to $\frac{a-b}{c} = \frac{2}{7}$, so the square root is $\sqrt{\frac{2}{7}}$, which matches option (A) $\frac{4}{7}$ only after noticing a misprint? Actually careful: the square root of $\frac{2}{7}$ is not $\frac{4}{7}$ — we must re‑evaluate: the expression inside the square root is $\frac{\sin(A-B)}{\sin(A+B)}$, and using known identities we get $\frac{a-b}{c} = \frac{2}{7}$, so the square root is $\sqrt{\frac{2}{7}}$, but none of the options match that. Wait — the problem likely intends $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a-b}{c}$, so the square root is $\sqrt{\frac{2}{7}}$, but option (A) is $\frac{4}{7}$ — that would be the square of $\frac{2}{7}$, not the root. Let’s check: Actually $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a^2-b^2}{c^2}$? No — the correct identity is $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a-b}{a+b}$? Let’s derive properly: final result is $\boxed{\frac{4}{7}}$ after squaring? No — the expression given is $\sqrt{\frac{\sin(A-B)}{\sin(A+B)}}$, and using $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a-b}{a+b}$? That gives $\frac{2}{8} = \frac14$, square root is $\frac12$, not an option. So the intended identity is $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a-b}{c}$? That gives $\frac{2}{7}$, square root $\sqrt{2/7}$ not an option. The only way to get $\frac{4}{7}$ is if the expression inside the root is $\frac{16}{49}$, i.e., $\frac{\sin(A-B)}{\sin(A+B)} = \frac{16}{49}$. Using $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a^2-b^2}{c^2}$? That gives $\frac{25-9}{49} = \frac{16}{49}$, so the square root is $\frac{4}{7}$. Yes — the correct identity is $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a^2-b^2}{c^2}$. So the answer is $\frac{4}{7}$, option (A).

    Using the law of sines and the sine subtraction/addition formulas, we find $\frac{\sin(A-B)}{\sin(A+B)} = \frac{a^2-b^2}{c^2} = \frac{16}{49}$, so its square root is $\frac{4}{7}$. The correct option is (A).

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    Concept & Intuition

    We have a triangle with all three sides given, but no angles directly. The expression involves $\sin(A-B)$ and $\sin(A+B)$. In any triangle, $A+B = \pi - C$, so $\sin(A+B) = \sin C$. That’s a nice simplification. For $\sin(A-B)$, we can use the law of sines to relate side lengths to sines of angles, then combine with the cosine rule or use known identities to express $\sin(A-B)$ in terms of sides. The key is to rewrite the ratio purely in terms of $a,b,c$.

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    Step‑by‑step solution

    1. Simplify the denominator In triangle $ABC$, $A+B+C = \pi$, so $A+B = \pi - C$. Hence

    $\sin(A+B) = \sin(\pi - C) = \sin C.$

    Therefore

    $\frac{\sin(A-B)}{\sin(A+B)} = \frac{\sin(A-B)}{\sin C}.$

    2. Express $\sin(A-B)$ using sides By the law of sines:

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

    So $\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, $\sin C = \frac{c}{2R}$.

    Now $\sin(A-B) = \sin A \cos B - \cos A \sin B$.

    We need $\cos A$ and $\cos B$. Use the law of cosines:

    $\cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{a^2 + c^2 - b^2}{2ac}.$

    3. Plug in the numbers Given $a=5$, $b=3$, $c=7$: $$$$

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11. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
    Q.In $\triangle ABC$, if $\dfrac{1}{a+b} + \dfrac{1}{c+a} = \dfrac{3}{a+b+c}$, then $\sin A =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\dfrac{4}{5}$

    ›Reveal solutionSolution

    The key idea is to simplify the given equation using the triangle side notation and then apply the Law of Cosines to find angle $A$. The result is $\sin A = \frac{\sqrt{3}}{2}$.

    The problem gives a relation involving the sides $a, b, c$ of triangle $ABC$. In any triangle, sides are opposite their respective angles: $a$ opposite $A$, $b$ opposite $B$, $c$ opposite $C$. The equation is purely algebraic in $a, b, c$, so we first simplify it to find a relationship between the sides. Once we have that, we can use the Law of Cosines to determine $\cos A$, and then $\sin A$.

    Let’s work through it step by step.

    1. Simplify the given equation. We have

    $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}.$

    Combine the left-hand side over a common denominator:

    $\frac{(c+a) + (a+b)}{(a+b)(c+a)} = \frac{2a + b + c}{(a+b)(c+a)}.$

    So the equation becomes

    $\frac{2a + b + c}{(a+b)(c+a)} = \frac{3}{a+b+c}.$

    2. Cross-multiply (all sides are positive, so no sign issues):

    $(2a + b + c)(a+b+c) = 3(a+b)(c+a).$

    3. Expand both sides. Left side:

    $(2a + b + c)(a+b+c) = 2a(a+b+c) + (b+c)(a+b+c).$

    Compute each part:

    $2a(a+b+c) = 2a^2 + 2ab + 2ac,$

    $(b+c)(a+b+c) = a(b+c) + (b+c)^2 = ab + ac + b^2 + 2bc + c^2.$

    Adding them:

    $2a^2 + 2ab + 2ac + ab + ac + b^2 + 2bc + c^2 = 2a^2 + 3ab + 3ac + b^2 + 2bc + c^2.$

    Right side:

    $3(a+b)(c+a) = 3[(a+b)(a+c)] = 3(a^2 + ac + ab + bc) = 3a^2 + 3ab + 3ac + 3bc.$

    4. Set them equal and simplify.

    $2a^2 + 3ab + 3ac + b^2 + 2bc + c^2 = 3a^2 + 3ab + 3ac + 3bc.$

    Cancel $3ab$ and $3ac$ from both sides:

    $2a^2 + b^2 + 2bc + c^2 = 3a^2 + 3bc.$

    Bring all terms to one side:

    $0 = 3a^2 + 3bc - 2a^2 - b^2 - 2bc - c^2 = a^2 + bc - b^2 - c^2.$

    So

    $a^2 + bc - b^2 - c^2 = 0 \quad \Rightarrow \quad a^2 = b^2 + c^2 - bc.$ …

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12. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.$\cot \dfrac{\pi}{16} \cdot \cot \dfrac{2\pi}{16} \cdot \cot \dfrac{3\pi}{16} \cdot \cot \dfrac{4\pi}{16} \cdot \cot \dfrac{5\pi}{16} \cdot \cot \dfrac{6\pi}{16} \cdot \cot \dfrac{7\pi}{16} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $2$

    ›Reveal solutionSolution

    The product of cotangents of complementary angles simplifies to 1; pairing each angle with its complement gives the result 1.

    The key insight is that cotangent has a beautiful symmetry: $\cot(\frac{\pi}{2} - \theta) = \tan \theta$, and $\cot \theta \cdot \tan \theta = 1$. So if we can pair each angle in the product with its complement (adding to $\frac{\pi}{2}$), the whole product collapses to 1 — except possibly for the middle term where the angle is exactly $\frac{\pi}{4}$, but $\cot \frac{\pi}{4} = 1$, so it doesn't break the pattern.

    Let's see this in action.

    1. Write the angles in radians:

       $\frac{\pi}{16}, \frac{2\pi}{16}, \frac{3\pi}{16}, \frac{4\pi}{16}, \frac{5\pi}{16}, \frac{6\pi}{16}, \frac{7\pi}{16}$.

    2. Notice that $\frac{\pi}{2} = \frac{8\pi}{16}$. So the complement of $\frac{k\pi}{16}$ is $\frac{8\pi}{16} - \frac{k\pi}{16} = \frac{(8-k)\pi}{16}$.

    3. Pair the terms:

       • $\frac{\pi}{16}$ pairs with $\frac{7\pi}{16}$ (since $1+7=8$)
       • $\frac{2\pi}{16}$ pairs with $\frac{6\pi}{16}$
       • $\frac{3\pi}{16}$ pairs with $\frac{5\pi}{16}$
       • $\frac{4\pi}{16}$ is its own complement (since $4+4=8$), i.e., $\frac{\pi}{4}$.

    4. For any pair $(\theta, \frac{\pi}{2} - \theta)$:

    $\cot \theta \cdot \cot\left(\frac{\pi}{2} - \theta\right) = \cot \theta \cdot \tan \theta = 1.$

    So each pair contributes a factor of 1.

    5. The middle term is $\cot \frac{4\pi}{16} = \cot \frac{\pi}{4} = 1$. …

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