Q.Integrate the function $\frac{\sin^8 x-\cos^8 x}{1-2\sin^2 x\cos^2 x}$

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\tan\alpha+2\tan2\alpha+4\tan4\alpha+8\cot8\alpha=$ (A) $\sin\alpha$ (B) $\cos\alpha$ (C) $\tan\alpha$ (D) $\cot\alpha$

   ›Reveal solutionSolution

   This is a telescoping trigonometric identity built from repeatedly doubling the angle; the sum collapses to $\cot\alpha$.

   Concept and Intuition

   The key building block is $\cot\theta - \tan\theta = 2\cot2\theta$ (from $\cot\theta-\tan\theta=\dfrac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{\cos2\theta}{\frac12\sin2\theta}=2\cot2\theta$), rearranged as $\cot\theta = \tan\theta+2\cot2\theta$. Applying this successively at $\theta=4\alpha, 2\alpha, \alpha$ turns each "cot of a doubled angle" term into a "tan plus cot of the next doubling," which cancels the corresponding tan term already present in the sum — a clean telescope.

   Step-by-Step Solution

   1. From $\cot4\alpha=\tan4\alpha+2\cot8\alpha$: $\;8\cot8\alpha = 4\cot4\alpha - 4\tan4\alpha$.
   2. Substitute into the expression: $\tan\alpha+2\tan2\alpha+4\tan4\alpha+(4\cot4\alpha-4\tan4\alpha) = \tan\alpha+2\tan2\alpha+4\cot4\alpha$.
   3. From $\cot2\alpha=\tan2\alpha+2\cot4\alpha$: $\;4\cot4\alpha=2\cot2\alpha-2\tan2\alpha$.
   4. Substitute: $\tan\alpha+2\tan2\alpha+(2\cot2\alpha-2\tan2\alpha)=\tan\alpha+2\cot2\alpha$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $f(x) = \dfrac{x}{1 - 3x^2} + \dfrac{x}{8}$, then $f(\tan 15^\circ) + f(\tan 20^\circ) =$ (A) $\dfrac{1 + \sqrt{3}}{8}$ (B) $\dfrac{3}{8}(3 + \sqrt{3})$ (C) $\dfrac{2 + \sqrt{3}}{8}$ (D) $\dfrac{3}{8}(1 + \sqrt{3})$

   ›Reveal solutionSolution

   The denominator $1-3x^2$ is designed to interact with the $60^\circ$ triple-angle tangent identity — plugging in $\tan15^\circ$ and $\tan20^\circ$ each collapses $f(x)$ to a clean closed form. The answer is (D).

   Concept and Intuition

   The triple angle formula $\tan3\theta = \dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}$ has exactly the denominator $1-3x^2$ that appears in $f(x)$. If $3\theta$ is a known angle (here $45^\circ$ for $\theta=15^\circ$, and $60^\circ$ for $\theta=20^\circ$), then $x=\tan\theta$ satisfies a specific cubic obtained by clearing denominators in the triple-angle identity — and that cubic is exactly what's needed to simplify $f(x)$ to a constant.

   Step-by-Step Solution

   1. For $\theta=15^\circ$: $3\theta=45^\circ$, so $\tan45^\circ=1=\dfrac{3x-x^3}{1-3x^2}$ where $x=\tan15^\circ$. This gives $1-3x^2=3x-x^3$, i.e. $x^3-3x^2-3x+1=0$.
   2. Suppose $f(x)=\dfrac{x}{1-3x^2}+\dfrac{x}{8}=k$ for a constant $k$. Then $\dfrac{x}{1-3x^2}=k-\dfrac{x}{8}=\dfrac{8k-x}{8}$, so $8x=(8k-x)(1-3x^2)$. Trying $k=\dfrac38$: $8x = (3-x)(1-3x^2) = 3-9x^2-x+3x^3$, i.e. $3x^3-9x^2-9x+3=0$, i.e. $x^3-3x^2-3x+1=0$ — exactly the equation from Step 1! So $f(\tan15^\circ)=\dfrac38$.
   3. For $\theta=20^\circ$: $3\theta=60^\circ$, so $\tan60^\circ=\sqrt3=\dfrac{3x-x^3}{1-3x^2}$ where $x=\tan20^\circ$. This gives $\sqrt3(1-3x^2)=3x-x^3$, i.e. $x^3-3\sqrt3x^2-3x+\sqrt3=0$. …

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3. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.If $\sin\theta + \csc\theta = 2$, then the value of $\sin^{10}\theta + \csc^{10}\theta =$ (A) $2$ (B) $2^{10}$ (C) $2^9$ (D) $2^8$

   ›Reveal solutionSolution

   The condition forces $\sin\theta=1$ exactly, so every power of $\sin\theta$ and $\csc\theta$ is just $1$, giving the sum $2$.

   Concept and Intuition

   For any positive real $s$, $s+\frac1s\ge2$ with equality only at $s=1$ (AM-GM). Since $\sin\theta\le1$, the given equation is only possible at the boundary case $\sin\theta=1$.

   Step-by-Step Solution

   1. Write the condition as $\sin\theta+\csc\theta=2 \Rightarrow \sin\theta+\frac{1}{\sin\theta}=2$.
   2. Multiply by $\sin\theta$: $\sin^2\theta-2\sin\theta+1=0$, i.e. $(\sin\theta-1)^2=0$. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\cos x + \cos^2 x = 1$, then $\sin^6 x + 3\sin^8 x + 3\sin^{10} x + \sin^{12} x =$ (A) $0$ (B) $1$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   This tests recognizing a binomial-cube pattern in disguise, combined with the given trig constraint. Answer: $1$.

   Concept and Intuition

   The expression $\sin^6x+3\sin^8x+3\sin^{10}x+\sin^{12}x$ has coefficients $1,3,3,1$ — exactly the binomial expansion of a cube. Recognizing $A^3 = A^3 + 3A^2B+3AB^2+B^3$ pattern (with $A=\sin^2x$... actually here it is $(\sin^2x)^3[1+3\sin^2x+3\sin^4x+\sin^6x]=(\sin^2 x)^3(1+\sin^2x)^3$) turns a messy sum into a single clean bracket, which the given condition on $\cos x$ then simplifies beautifully.

   Step-by-Step Solution

   1. From $\cos x+\cos^2x=1$: rearrange to $\cos x = 1-\cos^2x=\sin^2x$.
   2. Also rearranging the original equation directly: $\cos x + \cos^2 x = 1 \Rightarrow \cos x(1+\cos x)=1$.
   3. Factor the target expression:

   $\sin^6x+3\sin^8x+3\sin^{10}x+\sin^{12}x = \sin^6x\left(1+3\sin^2x+3\sin^4x+\sin^6x\right) = \sin^6x(1+\sin^2x)^3.$

   4. Write $\sin^6x = (\sin^2x)^3$, so the whole thing is $\left[\sin^2x(1+\sin^2x)\right]^3$. …

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5. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\sqrt{\sin^4 x + 4\cos^2 x} - \sqrt{\cos^4 x + 4\sin^2 x} =$ (A) $1-\cos 2x$ (B) $\tan 2x$ (C) $\sin 2x$ (D) $\cos 2x$

   ›Reveal solutionSolution

   Both radicands are perfect squares in disguise — the difference simplifies to (D) $\cos 2x$.

   Concept and Intuition

   Whenever you see $\sin^4x + k\cos^2x$ (or the cosine analogue), try converting the mixed powers to a single trig function using $\sin^2x+\cos^2x=1$, then look for a perfect-square pattern $(A-2)^2 = A^2-4A+4$.

   Step-by-Step Solution

   1. $\sin^4x+4\cos^2x = \sin^4x + 4(1-\sin^2x) = \sin^4x - 4\sin^2x + 4 = (\sin^2x-2)^2$.
   2. Since $0\le\sin^2x\le1$, we have $\sin^2x-2<0$, so $\sqrt{(\sin^2x-2)^2} = 2-\sin^2x$.
   3. Similarly, $\cos^4x+4\sin^2x = (\cos^2x-2)^2$, and $\sqrt{\cdots} = 2-\cos^2x$. …

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6. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.The value of $\tan\left(\dfrac{7\pi}{8}\right)$ is (A) $\sqrt{2}-1$ (B) $1-\sqrt{2}$ (C) $1+\sqrt{2}$ (D) $\dfrac{1}{1+\sqrt{2}}$

   ›Reveal solutionSolution

   Using $\tan(\pi-\theta)=-\tan\theta$ and the known value $\tan\frac{\pi}{8}=\sqrt2-1$, we get $\tan\frac{7\pi}{8}=1-\sqrt2$.

   Concept and Intuition

   Angles in the second quadrant can always be reduced to a first-quadrant reference angle using supplementary-angle identities; here $\frac{7\pi}{8}=\pi-\frac{\pi}{8}$.

   Step-by-Step Solution

   1. Write $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$.
   2. Use $\tan(\pi-\theta) = -\tan\theta$, so $\tan\frac{7\pi}{8} = -\tan\frac{\pi}{8}$.
   3. Recall (or derive from half-angle formula with $\theta=\pi/4$): $\tan\frac{\pi}{8} = \tan22.5^\circ = \sqrt2-1$. …

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7. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} =$ (A) $2\sec x$ (B) $2\csc x$ (C) $\tan 2x$ (D) $\sin 2x$

   ›Reveal solutionSolution

   Combining the two fractions and using the Pythagorean identity collapses everything to $2\csc x$.

   Concept and Intuition

   Whenever you see a sum of a fraction and its "flipped" reciprocal-like partner, combining over a

   common denominator and using $\sin^2x+\cos^2x=1$ is almost always the fastest route — the

   $(1+\cos x)^2$ expansion conveniently reintroduces $\sin^2x+\cos^2x$, letting everything collapse.

   Step-by-Step Solution

   1. Write the sum with common denominator $\sin x(1+\cos x)$:

   $\frac{\sin x}{1+\cos x}+\frac{1+\cos x}{\sin x} = \frac{\sin^2 x + (1+\cos x)^2}{\sin x(1+\cos x)}.$

   2. Expand the numerator: $\sin^2x + 1+2\cos x+\cos^2x = (\sin^2x+\cos^2x)+1+2\cos x = 1+1+2\cos x = 2+2\cos x$. …

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8. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\dfrac{1}{1+\sin\theta}+\dfrac{1}{1-\sin\theta} =$ (A) $2\cos^2\theta$ (B) $-2\cos^2\theta$ (C) $2\tan^2\theta$ (D) $2\sec^2\theta$

   ›Reveal solutionSolution

   This tests combining two fractions over a common denominator and simplifying using the Pythagorean identity $1-\sin^2\theta=\cos^2\theta$. Answer: $2\sec^2\theta$.

   Concept and Intuition

   Adding fractions with denominators that are conjugates of each other, $(1+\sin\theta)$ and $(1-\sin\theta)$, produces a difference-of-squares denominator, which simplifies beautifully using the fundamental identity $\sin^2\theta+\cos^2\theta=1$.

   Step-by-Step Solution

   1. Find a common denominator: $\dfrac{1}{1+\sin\theta}+\dfrac{1}{1-\sin\theta} = \dfrac{(1-\sin\theta)+(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}$.
   2. Numerator simplifies: $(1-\sin\theta)+(1+\sin\theta) = 2$.
   3. Denominator is a difference of squares: $(1+\sin\theta)(1-\sin\theta) = 1-\sin^2\theta$.
   4. Use $1-\sin^2\theta=\cos^2\theta$: denominator becomes $\cos^2\theta$. …

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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\sin x > 0$ and $\sin^2 x + \sin^4 x + 2\sin^3 x = 1$, then $\cos^{16}x + \cos^8 x + 4(\cos^{14}x + \cos^{10}x) + 6\cos^{12}x + 1 =$ (A) 1 (B) 2 (C) 3 (D) 4

   ›Reveal solutionSolution

   The given constraint pins $\cos^2x$ down to a root of $u^2+u=1$; the target expression collapses to $(u^2+u)^4+1=2$.

   Concept and Intuition

   The expression $\sin^2x+\sin^4x+2\sin^3x$ is a disguised perfect-square-difference: grouping as $(\sin^2x+\sin x)^2-\sin^2x$ reveals it factors via difference of squares, which is much faster than brute-force solving a quartic. The pay-off expression in $\cos x$, with its 1-4-6-4-1-looking coefficients, is recognizable as the binomial expansion $(c^2+c)^4$ plus the leftover $+1$ — once you see that pattern, you just need the value of $c^2+c$ where $c=\cos^2x$.

   Step-by-Step Solution

   1. Let $t=\sin x$. The equation is $t^2+t^4+2t^3=1$, i.e. $t^4+2t^3+t^2-1=0$.
   2. Notice $t^4+2t^3+t^2=(t^2+t)^2$, so the equation is $(t^2+t)^2-1=0=(t^2+t-1)(t^2+t+1)$.
   3. $t^2+t+1=(t+\tfrac12)^2+\tfrac34>0$ always, so we need $t^2+t-1=0$, giving $t=\dfrac{-1\pm\sqrt5}{2}$. Since $\sin x>0$, take $t=\dfrac{\sqrt5-1}{2}\approx0.618$.
   4. From $t^2+t-1=0$, $t^2=1-t$, so $\sin^2x=1-\sin x$, hence $\cos^2x=1-\sin^2x=\sin x=t$. So $\cos^2x$ itself equals $t$, and thus $c:=\cos^2x$ also satisfies $c^2+c=1$ (same equation, since $c=t$). …

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10. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
    Q.$(1 - \tan 348^\circ)(1 + \cot 417^\circ) =$ (A) $3\sqrt{3}$ (B) 2 (C) $\dfrac{2}{\sqrt{3}}$ (D) 1

    ›Reveal solutionSolution

    Reduce both angles to the first quadrant and recognize the classic $(1+\tan A)(1+\tan B)=2$ identity for $A+B=45^\circ$. Answer: (B).

    Concept and Intuition

    When two angles sum to $45^\circ$, their tangents satisfy $\tan A+\tan B+\tan A\tan B=1$ (from the tangent addition formula with $\tan45^\circ=1$), which makes $(1+\tan A)(1+\tan B)=2$ — a well-known shortcut identity.

    Step-by-Step Solution

    1. $\tan348^\circ = \tan(360^\circ-12^\circ) = -\tan12^\circ$, so $1-\tan348^\circ = 1-(-\tan12^\circ) = 1+\tan12^\circ$.
    2. $\cot417^\circ = \cot(417^\circ-360^\circ) = \cot57^\circ = \tan(90^\circ-57^\circ) = \tan33^\circ$, so $1+\cot417^\circ = 1+\tan33^\circ$.
    3. Expression becomes $(1+\tan12^\circ)(1+\tan33^\circ)$. …

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11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.$[1+\sec2\theta][1+\sec4\theta]=$ (A) $\tan\theta\tan4\theta$ (B) $4\cot\theta\tan4\theta$ (C) $\cot\theta\tan4\theta$ (D) $4\tan\theta\tan4\theta$

    ›Reveal solutionSolution

    Rewrite each $1+\sec(\cdot)$ factor using $1+\cos x=2\cos^2(x/2)$, then simplify the ratio using the double-angle identity for sine. Answer: $\cot\theta\tan4\theta$.

    Concept and Intuition

    The key trick for $1+\sec x$ expressions is converting to $\dfrac{1+\cos x}{\cos x}=\dfrac{2\cos^2(x/2)}{\cos x}$, which turns a sum into a clean ratio of cosines at half the angle. Doing this twice (for $2\theta$ and $4\theta$) and multiplying collapses most of the cosines, leaving a tangent-cotangent ratio via $\sin4\theta=4\sin\theta\cos\theta\cos2\theta$.

    Step-by-Step Solution

    1. $1+\sec2\theta=1+\dfrac{1}{\cos2\theta}=\dfrac{\cos2\theta+1}{\cos2\theta}=\dfrac{2\cos^2\theta}{\cos2\theta}$ (using $1+\cos2\theta=2\cos^2\theta$).
    2. Similarly, $1+\sec4\theta=\dfrac{2\cos^2(2\theta)}{\cos4\theta}$ (using $1+\cos4\theta=2\cos^2(2\theta)$).
    3. Multiply: $[1+\sec2\theta][1+\sec4\theta]=\dfrac{2\cos^2\theta}{\cos2\theta}\cdot\dfrac{2\cos^2(2\theta)}{\cos4\theta}=\dfrac{4\cos^2\theta\cos2\theta}{\cos4\theta}$ (one factor of $\cos2\theta$ cancels). …

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12. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.$\cos\dfrac{\pi}{2^2}\cdot\cos\dfrac{\pi}{2^3}\cdot\cos\dfrac{\pi}{2^4}\cdots\cos\dfrac{\pi}{2^{10}}=$ (A) $\dfrac{\sin\left(\frac{\pi}{2^{10}}\right)}{512}$ (B) $\dfrac{\csc\left(\frac{\pi}{2^{10}}\right)}{512}$ (C) $\dfrac{\sin\left(\frac{\pi}{2^{10}}\right)}{1024}$ (D) $\dfrac{\csc\left(\frac{\pi}{2^{10}}\right)}{1024}$

    ›Reveal solutionSolution

    The telescoping identity $\cos\theta=\sin2\theta/(2\sin\theta)$ collapses the whole product of cosines into $\dfrac{\csc(\pi/2^{10})}{512}$.

    Concept and Intuition

    Products of cosines of successively halved angles are classic telescoping problems, using $\sin2\theta=2\sin\theta\cos\theta \Rightarrow \cos\theta=\dfrac{\sin2\theta}{2\sin\theta}$. Writing each cosine this way causes almost all intermediate sine terms to cancel, leaving only the first and last.

    Step-by-Step Solution

    1. Write $\cos\dfrac{\pi}{2^k} = \dfrac{\sin(\pi/2^{k-1})}{2\sin(\pi/2^k)}$ for each $k=2,\ldots,10$.
    2. Multiply all nine such fractions ($k=2$ to $10$): the numerator of each term is $\sin(\pi/2^{k-1})$, which is exactly the denominator's sine argument from the previous term's neighbor — this telescopes, cancelling every intermediate sine.
    3. What survives: numerator $\sin(\pi/2^1)=\sin(\pi/2)=1$ (from the very first term, $k=2$), and denominator $2^9\sin(\pi/2^{10})$ (the $2^9$ from nine factors of $2$, and the final uncancelled sine from the last term, $k=10$). …

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