Q.Evaluate: $\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx$

Showing the 12 most recent of 40 on this concept.

1. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.$\sin 21^\circ \cos 9^\circ - \cos 84^\circ \cos 6^\circ =$ (A) 1 (B) $\dfrac{1}{4}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{3}{2}$

   ›Reveal solutionSolution

   Convert $\cos84^\circ$ to $\sin6^\circ$ and expand both products using sum-to-product identities; the $\sin12^\circ$ terms cancel exactly, leaving $\dfrac14$.

   Concept and Intuition

   When an expression mixes cosines and sines of complementary-looking angles ($84^\circ=90^\circ-6^\circ$), converting everything to a common trig function often reveals hidden cancellation via product-to-sum formulas.

   Step-by-Step Solution

   1. Note $\cos84^\circ = \sin(90^\circ-84^\circ)=\sin6^\circ$.
   2. So $\cos84^\circ\cos6^\circ = \sin6^\circ\cos6^\circ = \dfrac12\sin12^\circ$ (double-angle identity). …

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2. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If $A+B+C=\dfrac{\pi}{4}$, then $\sin 4A+\sin 4B+\sin 4C=$ (A) $4\cos 2A\cos 2B\cos 2C$ (B) $4\sin 2A\sin 2B\sin 2C$ (C) $1+4\sin 2A\sin 2B\sin 2C$ (D) $1+4\cos 2A\cos 2B\cos 2C$

   ›Reveal solutionSolution

   This tests recognizing a disguised "triangle-angle" identity after a substitution. Answer: $\sin4A+\sin4B+\sin4C=4\cos2A\cos2B\cos2C$.

   Concept and Intuition

   The classic identity "if $X+Y+Z=\pi$ then $\sin X+\sin Y+\sin Z=4\cos\frac X2\cos\frac Y2\cos\frac Z2$" is usually stated for a triangle's angles, but it's really just a trigonometric identity that holds for any three angles summing to $\pi$ — it doesn't care where the angles came from. Here $A+B+C=\pi/4$ looks unrelated at first, but multiplying every angle by 4 turns the condition into exactly $X+Y+Z=\pi$ with $X=4A$ etc., unlocking the identity immediately.

   Step-by-Step Solution

   1. Given $A+B+C=\dfrac{\pi}{4}$. Multiply by 4: $4A+4B+4C=\pi$.
   2. Let $X=4A,\ Y=4B,\ Z=4C$, so $X+Y+Z=\pi$.
   3. Apply the standard identity (valid whenever three angles sum to $\pi$): $\sin X+\sin Y+\sin Z=4\cos\dfrac X2\cos\dfrac Y2\cos\dfrac Z2$.
   4. Here $\dfrac X2=2A,\ \dfrac Y2=2B,\ \dfrac Z2=2C$.
   5. So $\sin4A+\sin4B+\sin4C=4\cos2A\cos2B\cos2C$.

   Common Mistakes …

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3. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.If $\cos^3 x \sin 4x = \sum_{r=0}^{n} a_r \sin rx$ $\forall x \in \mathbb{R}$, then $a_3+a_5 : a_1+a_7 =$ (A) $1:3$ (B) $1:1$ (C) $2:1$ (D) $3:1$

   ›Reveal solutionSolution

   Expanding $\cos^3x\sin4x$ as a sum of sines via product-to-sum identities gives coefficients $a_1=a_7=\frac18$, $a_3=a_5=\frac38$, so the requested ratio is $3:1$.

   Concept and Intuition

   $\cos^3x$ reduces to a linear combination of $\cos x$ and $\cos3x$ (a standard multiple-angle reduction), turning the product into a sum of two simpler $\cos\cdot\sin$ products, each of which splits into two sine terms via product-to-sum formulas.

   Step-by-Step Solution

   1. $\cos^3x=\dfrac{3\cos x+\cos3x}{4}$ (from $\cos3x=4\cos^3x-3\cos x$).
   2. $\cos^3x\sin4x=\dfrac34\cos x\sin4x+\dfrac14\cos3x\sin4x$.
   3. $\cos x\sin4x=\dfrac12[\sin(4x+x)+\sin(4x-x)]=\dfrac12(\sin5x+\sin3x)$.
   4. $\cos3x\sin4x=\dfrac12[\sin(4x+3x)+\sin(4x-3x)]=\dfrac12(\sin7x+\sin x)$.
   5. Combine: $\cos^3x\sin4x=\dfrac34\cdot\dfrac12(\sin5x+\sin3x)+\dfrac14\cdot\dfrac12(\sin7x+\sin x)=\dfrac38\sin5x+\dfrac38\sin3x+\dfrac18\sin7x+\dfrac18\sin x$. …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The general solution of the equation $\sin x - 3\sin 2x + \sin 3x = \cos x - 3\cos 2x + \cos 3x$ is (A) $n\pi + \dfrac{\pi}{8}, n \in \mathbb{Z}$ (B) $\dfrac{n\pi}{2} + \dfrac{\pi}{8}, n \in \mathbb{Z}$ (C) $(-1)^n\dfrac{n\pi}{2} + \dfrac{\pi}{8}, n \in \mathbb{Z}$ (D) $2n\pi + \cos^{-1}\dfrac{3}{2}, n \in \mathbb{Z}$

   ›Reveal solutionSolution

   A sum-to-product regrouping turns the equation into $(2\cos x-3)(\sin2x-\cos2x)=0$; the first factor never vanishes, so the general solution is $x=\frac{n\pi}{2}+\frac{\pi}{8}$.

   Concept and Intuition

   When an equation mixes $x$, $2x$, $3x$ terms, grouping the outer terms ($x$ and $3x$) via sum-to-product often exposes a common factor with the middle term — turning a transcendental-looking equation into a simple factorable one.

   Step-by-Step Solution

   1. $\sin x+\sin 3x = 2\sin 2x\cos x$ and $\cos x + \cos 3x = 2\cos 2x\cos x$ (sum-to-product, since the mean angle is $2x$ and half-difference is $x$).
   2. LHS of the given equation $=2\sin2x\cos x - 3\sin2x = \sin2x(2\cos x - 3)$.
   3. RHS $=2\cos2x\cos x - 3\cos2x = \cos2x(2\cos x-3)$.
   4. Equation: $\sin2x(2\cos x-3)=\cos2x(2\cos x-3)$, i.e. $(2\cos x-3)(\sin2x-\cos2x)=0$.
   5. $2\cos x = 3$ has no real solution (cosine is bounded by 1), so we must have $\sin2x=\cos2x$, i.e. $\tan2x=1$. …

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5. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\cos6^{\circ}\sin24^{\circ}\cos72^{\circ}=$ (A) $-\dfrac{1}{8}$ (B) $-\dfrac{1}{4}$ (C) $\dfrac{1}{8}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   This is a product of three trig values at special angles that evaluates to a clean constant, $1/8$.

   Concept and Intuition

   $\cos72°=\sin18°$ and this product of three cosine/sine values at angles related by factors tied to $18°/36°$ is a disguised instance of the well-known identity $\cos\theta\cos(60°-\theta)\cos(60°+\theta)=\tfrac14\cos3\theta$-family of results; here it is most reliably confirmed by direct exact evaluation.

   Step-by-Step Solution

   1. Rewrite $\sin24°=\cos66°$, so the product is $\cos6°\cos66°\cos72°$.
   2. Use $\cos A\cos B = \tfrac12[\cos(A-B)+\cos(A+B)]$ on $\cos6°\cos66°$: $=\tfrac12[\cos60°+\cos72°]=\tfrac12\left[\tfrac12+\cos72°\right]$.
   3. So the full product is $\tfrac12\left[\tfrac12+\cos72°\right]\cos72° = \tfrac14\cos72° + \tfrac12\cos^272°$.
   4. Using the exact value $\cos72°=\dfrac{\sqrt5-1}{4}$: $\cos^272°=\dfrac{(\sqrt5-1)^2}{16}=\dfrac{6-2\sqrt5}{16}=\dfrac{3-\sqrt5}{8}$. …

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6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $x \neq (2n+1)\dfrac{\pi}{4}$, then the general solution of $\cos x + \cos 3x = \sin x + \sin 3x$ is (A) $n\pi + \dfrac{\pi}{8}$ (B) $n\pi \pm \dfrac{\pi}{8}$ (C) $\dfrac{n\pi}{2} \pm \dfrac{\pi}{8}$ (D) $\dfrac{n\pi}{2} + \dfrac{\pi}{8}$

   ›Reveal solutionSolution

   Convert both sides to a product form using sum-to-product identities, factor out $\cos x$, and solve $\tan2x=1$ for the general solution — the given domain restriction is exactly what makes this branch well-defined. Answer: $x=\dfrac{n\pi}{2}+\dfrac\pi8$.

   Concept and Intuition

   Sum-to-product identities turn a sum of cosines/sines into a product, which is powerful for solving trig equations because a product equal to zero splits into simpler independent equations. Here both sums share the common factor $\cos x$, and the interesting equation reduces to a basic $\tan2x=1$.

   Step-by-Step Solution

   1. $\cos x+\cos3x=2\cos\!\left(\dfrac{x+3x}{2}\right)\cos\!\left(\dfrac{3x-x}{2}\right)=2\cos2x\cos x$.
   2. $\sin x+\sin3x=2\sin\!\left(\dfrac{x+3x}{2}\right)\cos\!\left(\dfrac{3x-x}{2}\right)=2\sin2x\cos x$.
   3. The equation $\cos x+\cos3x=\sin x+\sin3x$ becomes $2\cos2x\cos x=2\sin2x\cos x$, i.e. $2\cos x(\cos2x-\sin2x)=0$.
   4. The condition $x\ne(2n+1)\pi/4$ is precisely the condition $\cos2x\ne0$, which is exactly what's needed to safely write $\tan2x=\dfrac{\sin2x}{\cos2x}$ from $\cos2x=\sin2x$ — signalling that the intended branch to solve is $\cos2x-\sin2x=0$. …

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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\sum_{k=0}^{12} \dfrac{1}{\sin\left((k+1)\dfrac{\pi}{6}+\dfrac{\pi}{4}\right)\sin\left(\dfrac{k\pi}{6}+\dfrac{\pi}{4}\right)} =$ (A) $2(\sqrt{3}+1)$ (B) $2(3-\sqrt{3})$ (C) $2(2-\sqrt{3})$ (D) $2(\sqrt{3}-1)$

   ›Reveal solutionSolution

   This is a telescoping trigonometric sum using the cotangent-difference identity. The answer is (D).

   Concept and Intuition

   Whenever consecutive terms in a sum $\dfrac{1}{\sin A_k \sin A_{k-1}}$ have a constant angular gap $d=A_k-A_{k-1}$, the identity $\dfrac{\sin(A-B)}{\sin A\sin B}=\cot B-\cot A$ turns each term into a difference of cotangents, so the whole sum telescopes to just the first and last cotangent.

   Step-by-Step Solution

   1. Let $A_k = k\dfrac{\pi}{6}+\dfrac{\pi}{4}$ for $k=0,\dots,13$. Each term is $\dfrac{1}{\sin A_{k+1}\sin A_k}$, and $A_{k+1}-A_k=\pi/6$ always.
   2. Using $\dfrac{\sin(\pi/6)}{\sin A_{k+1}\sin A_k} = \cot A_k - \cot A_{k+1}$, each term equals $\dfrac{1}{\sin(\pi/6)}(\cot A_k-\cot A_{k+1}) = 2(\cot A_k - \cot A_{k+1})$.
   3. Summing $k=0$ to $12$ telescopes: $\sum = 2(\cot A_0 - \cot A_{13})$.
   4. $A_0=\pi/4$, so $\cot A_0=1$. …

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8. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The value of $\sin(\frac{5\pi}{24}) \cdot \cos(\frac{\pi}{24})$ is (A) $\frac{1+\sqrt2}{4}$ (B) $1+\sqrt2$ (C) $\frac{1-\sqrt2}{4}$ (D) $1-\sqrt2$

   ›Reveal solutionSolution

   A direct application of the product-to-sum formula turns the awkward angles $5\pi/24$ and $\pi/24$

   into the familiar $\pi/4$ and $\pi/6$; the value is $\dfrac{1+\sqrt2}{4}$.

   Concept and Intuition

   Products of sine and cosine at "ugly" angles often simplify beautifully once you notice that their

   sum and difference are standard angles. Here $\frac{5\pi}{24}+\frac{\pi}{24}=\frac{6\pi}{24}=\frac{\pi}{4}$

   and $\frac{5\pi}{24}-\frac{\pi}{24}=\frac{4\pi}{24}=\frac{\pi}{6}$ — both angles whose sine we know exactly.

   This is exactly the situation the product-to-sum identity is built for.

   Step-by-Step Solution

   1. Recall $\sin A\cos B=\dfrac12\left[\sin(A+B)+\sin(A-B)\right]$.
   2. Here $A=\dfrac{5\pi}{24}$, $B=\dfrac{\pi}{24}$, so $A+B=\dfrac{\pi}{4}$, $A-B=\dfrac{\pi}{6}$.
   3. $\sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}$ and $\sin\dfrac{\pi}{6}=\dfrac12$. …

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9. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Statement - I: If $x \in \left(0, \dfrac{\pi}{2}\right)$ and $\cos 3x + \cos x = \cos 2x$, then $x = \dfrac{\pi}{4}$ or $\dfrac{\pi}{3}$ Statement - II: If $\sin x \sin 2x = \cos x \cos 2x - 1$, then $x = \dfrac{n\pi}{3}, n \in Z$ (A) I is true and II is true (B) I is false and II is true (C) I is true and II is false (D) I is false and II is false

   ›Reveal solutionSolution

   This checks two trig-equation general solutions, one true and one deliberately over-stated. Answer: I true, II false.

   Concept and Intuition

   Sum-to-product identities collapse $\cos3x+\cos x$ and combine-to-single-angle identities collapse $\cos x\cos2x-\sin x\sin2x$ into $\cos3x$. The key discipline is getting the general solution exactly right — $\cos\theta=1$ gives $\theta=2n\pi$ (even multiples), not just any integer multiple of the reduced angle.

   Step-by-Step Solution

   Statement I:

   1. $\cos3x+\cos x = 2\cos\left(\dfrac{3x+x}{2}\right)\cos\left(\dfrac{3x-x}{2}\right) = 2\cos2x\cos x$.
   2. Equation becomes $2\cos2x\cos x=\cos2x \Rightarrow \cos2x(2\cos x-1)=0$.
   3. Case $\cos2x=0$: for $x\in(0,\pi/2)$, $2x\in(0,\pi)$, so $2x=\pi/2 \Rightarrow x=\pi/4$.
   4. Case $\cos x=1/2$: $x=\pi/3$ (within range).
   5. So $x=\pi/4$ or $\pi/3$ — Statement I is true.

   Statement II:

   6. $\sin x\sin2x=\cos x\cos2x-1 \Rightarrow \cos x\cos2x-\sin x\sin2x=1 \Rightarrow \cos(x+2x)=1 \Rightarrow \cos3x=1$.

   7. General solution of $\cos\theta=1$ is $\theta=2n\pi$ ($n\in\mathbb{Z}$), so $3x=2n\pi \Rightarrow x=\dfrac{2n\pi}{3}$. …

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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $P+Q+R=\dfrac{\pi}{4}$, then $\cos\left(\dfrac{\pi}{8} - P\right) + \cos\left(\dfrac{\pi}{8} - Q\right) + \cos\left(\dfrac{\pi}{8} - R\right) =$ (A) $4\cos\dfrac{P}{2}\cos\dfrac{Q}{2}\cos\dfrac{R}{2} - \cos\dfrac{\pi}{8}$ (B) $4\cos\dfrac{P}{2}\cos\dfrac{Q}{2}\sin\dfrac{R}{2} + \cos\dfrac{\pi}{8}$ (C) $4\sin\dfrac{P}{2}\sin\dfrac{Q}{2}\sin\dfrac{R}{2} - \cos\dfrac{\pi}{8}$ (D) $4\sin\dfrac{P}{2}\cos\dfrac{Q}{2}\sin\dfrac{R}{2} + \cos\dfrac{\pi}{8}$

    ›Reveal solutionSolution

    This tests the sum-to-product technique for three angles constrained by $P+Q+R=\pi/4$; the answer is $4\cos\frac{P}{2}\cos\frac{Q}{2}\cos\frac{R}{2}-\cos\frac{\pi}{8}$.

    Concept and Intuition

    Whenever three angles are linked by a linear constraint like $P+Q+R=$ constant, the standard move is to combine two of the cosine terms via sum-to-product so that the constraint eliminates one variable, then use product-to-sum again to fold the remaining pieces into a symmetric product. This mirrors the classical identity for a triangle ($A+B+C=\pi \Rightarrow \cos A+\cos B+\cos C = 1+4\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$), just with a different target constant ($\pi/4$ instead of $\pi$).

    Step-by-Step Solution

    1. Combine the first two terms using $\cos C+\cos D = 2\cos\frac{C+D}{2}\cos\frac{C-D}{2}$ with $C=\frac{\pi}{8}-P,\ D=\frac{\pi}{8}-Q$:

    $\cos\left(\tfrac{\pi}{8}-P\right)+\cos\left(\tfrac{\pi}{8}-Q\right)=2\cos\left(\tfrac{\pi}{8}-\tfrac{P+Q}{2}\right)\cos\left(\tfrac{Q-P}{2}\right).$

    2. Since $P+Q=\frac{\pi}{4}-R$, we get $\frac{\pi}{8}-\frac{P+Q}{2}=\frac{\pi}{8}-\frac{\pi}{8}+\frac{R}{2}=\frac{R}{2}$. So the pair sums to $2\cos\frac{R}{2}\cos\left(\frac{Q-P}{2}\right)$.
    3. Use product-to-sum on $\cos\frac{P}{2}\cos\frac{Q}{2}$: $2\cos\frac{P}{2}\cos\frac{Q}{2}=\cos\left(\frac{Q-P}{2}\right)+\cos\left(\frac{\pi}{8}-\frac{R}{2}\right)$ (using $\frac{P+Q}{2}=\frac{\pi}{8}-\frac{R}{2}$ again). So $\cos\left(\frac{Q-P}{2}\right)=2\cos\frac{P}{2}\cos\frac{Q}{2}-\cos\left(\frac{\pi}{8}-\frac{R}{2}\right)$.
    4. Substitute: $2\cos\frac{R}{2}\cos\left(\frac{Q-P}{2}\right)=4\cos\frac{P}{2}\cos\frac{Q}{2}\cos\frac{R}{2}-2\cos\frac{R}{2}\cos\left(\frac{\pi}{8}-\frac{R}{2}\right)$.
    5. Expand the last product using $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ with $A=\frac{R}{2},B=\frac{\pi}{8}-\frac{R}{2}$: it equals $\cos\left(\frac{\pi}{8}-R\right)+\cos\frac{\pi}{8}$. …

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11. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.If $A + B + C = \dfrac{3\pi}{2}$, then $\cos 2A + \cos 2B + \cos 2C =$ (A) $1 - 4\sin A \sin B \sin C$ (B) $1 + 4\sin A \sin B \sin C$ (C) $1 - 2\sin A \sin B \sin C$ (D) $1 + 2\sin A \sin B \sin C$

    ›Reveal solutionSolution

    For $A+B+C=\frac{3\pi}{2}$, the identity $\cos2A+\cos2B+\cos2C = 1-4\sin A\sin B\sin C$ holds (the sign pattern differs from the more famous $A+B+C=\pi$ version).

    Concept and Intuition

    Whenever angles are constrained to sum to a fixed value, sum-to-product formulas convert the sum of cosines of double angles into a product involving the sines of the individual angles — this is the standard trick behind many triangle-identity MCQs.

    Step-by-Step Solution

    1. Combine the first two terms: $\cos2A+\cos2B = 2\cos(A+B)\cos(A-B)$.
    2. Since $A+B+C=\frac{3\pi}{2}$, we have $A+B = \frac{3\pi}{2}-C$, so $\cos(A+B)=\cos\left(\frac{3\pi}{2}-C\right) = -\sin C$ (using $\cos(\frac{3\pi}{2}-x) = -\sin x$).
    3. So $\cos2A+\cos2B = -2\sin C\cos(A-B)$.
    4. Add $\cos2C = 1-2\sin^2C$: total $= 1-2\sin^2C - 2\sin C\cos(A-B) = 1 - 2\sin C\left[\sin C+\cos(A-B)\right]$.
    5. Since $C = \frac{3\pi}{2}-(A+B)$, $\sin C = \sin\left(\frac{3\pi}{2}-(A+B)\right) = -\cos(A+B)$. So $\sin C+\cos(A-B) = \cos(A-B)-\cos(A+B) = 2\sin A\sin B$. …

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12. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.If $A+B+C+D=2\pi$, then $\sin A+\sin B+\sin C+\sin D=$ (A) $4\sin\left(\dfrac{A+B}{4}\right)\sin\left(\dfrac{A+C}{4}\right)\sin\left(\dfrac{A+D}{4}\right)$ (B) $4\sin\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A+C}{4}\right)\cos\left(\dfrac{A+D}{4}\right)$ (C) $4\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A+C}{2}\right)\sin\left(\dfrac{A+D}{2}\right)$ (D) $4\sin\left(\dfrac{A+B}{2}\right)\sin\left(\dfrac{A+C}{4}\right)\sin\left(\dfrac{A+D}{4}\right)$

    ›Reveal solutionSolution

    This is a standard four-angle sum-to-product identity valid whenever $A+B+C+D=2\pi$. Answer: $4\sin\frac{A+B}2\sin\frac{A+C}2\sin\frac{A+D}2$.

    Concept and Intuition

    Pairing $(\sin A+\sin B)$ and $(\sin C+\sin D)$ separately, then using the constraint $A+B+C+D=2\pi$ to relate the two half-sum angles as supplementary, lets a second product-to-sum step factor the whole thing into three sine factors — a symmetric pattern.

    Step-by-Step Solution

    1. $\sin A+\sin B=2\sin\dfrac{A+B}2\cos\dfrac{A-B}2$ and $\sin C+\sin D=2\sin\dfrac{C+D}2\cos\dfrac{C-D}2$.
    2. Since $A+B+C+D=2\pi$, $\dfrac{C+D}2=\pi-\dfrac{A+B}2$, so $\sin\dfrac{C+D}2=\sin\dfrac{A+B}2$.
    3. So the sum $=2\sin\dfrac{A+B}2\left[\cos\dfrac{A-B}2+\cos\dfrac{C-D}2\right]$.
    4. Apply sum-to-product on the bracket: $\cos\dfrac{A-B}2+\cos\dfrac{C-D}2=2\cos\dfrac{A-B+C-D}4\cos\dfrac{A-B-C+D}4$.
    5. Using $A+B+C+D=2\pi$ one can rewrite $\dfrac{A-B+C-D}4=\dfrac{A+C}2-\dfrac{\pi}2+\dots$ type manipulations (or equivalently verify numerically) to reduce the whole expression to the symmetric form $4\sin\dfrac{A+B}2\sin\dfrac{A+C}2\sin\dfrac{A+D}2$.
    6. Check with a concrete case: $A=B=C=D=\pi/2$ (sum $=2\pi$): LHS $=4\sin90°=4$. Option (C): $4\sin90°\sin90°\sin90°=4$. ✓ Options (A),(B),(D) all give values $\ne4$ here, eliminating them. …

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