Q.Integrate the following function: $\frac{\sin^2 x}{1 + \cos x}$

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If A and B are positive acute angles satisfying $3\cos^2 A + 2\cos^2 B = 4$ and $\dfrac{3\sin A}{\sin B} = \dfrac{2\cos B}{\cos A}$, then $A + 2B =$ (A) $30^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$

   ›Reveal solutionSolution

   This tests combining two trig constraint equations using double-angle identities to pin down a specific angle sum. The result is $A+2B=90^\circ$.

   Concept and Intuition

   When given two independent trigonometric equations in two unknown angles, converting everything into double-angle form (via $\cos2\theta=1-2\sin^2\theta$ and $\sin2\theta=2\sin\theta\cos\theta$) turns the problem into solving simultaneous equations in $\cos2A$ and $\cos2B$ (or equivalently $\sin2A,\sin2B$), using the Pythagorean identity to close the system.

   Step-by-Step Solution

   1. Rewrite $3\cos^2A+2\cos^2B=4$ using $\cos^2\theta=1-\sin^2\theta$: $3(1-\sin^2A)+2(1-\sin^2B)=4 \implies 5-3\sin^2A-2\sin^2B=4 \implies 3\sin^2A+2\sin^2B=1$.
   2. So $3\sin^2A = 1-2\sin^2B=\cos2B$. Call this Eq I: $\cos2B = 3\sin^2A = \dfrac{3(1-\cos2A)}{2}$.
   3. From the second given equation $3\sin A\cos A = 2\sin B\cos B$, i.e. $\dfrac{3}{2}\sin2A=\sin2B$. Call this Eq II: $\sin2B=\dfrac{3}{2}\sin2A$.
   4. Substitute Eq I and Eq II into $\sin^22B+\cos^22B=1$:

   $\left(\dfrac{3}{2}\sin2A\right)^2+\left(\dfrac{3(1-\cos2A)}{2}\right)^2=1$

   $\dfrac{9}{4}\left[\sin^22A+(1-\cos2A)^2\right]=1$

   5. Expand: $\sin^22A+(1-\cos2A)^2 = \sin^22A+1-2\cos2A+\cos^22A = 2-2\cos2A$. So $\dfrac{9}{4}\cdot2(1-\cos2A)=1 \implies 1-\cos2A=\dfrac{2}{9} \implies \cos2A=\dfrac{7}{9}$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\sin(\pi\cos\theta) = \cos(\pi\sin\theta)$, then $\sin 2\theta =$ (A) $\pm\dfrac{3}{4}$ (B) $\pm\dfrac{1}{2}$ (C) $\pm\dfrac{1}{\sqrt{3}}$ (D) $\pm\sqrt{2}$

   ›Reveal solutionSolution

   Convert $\cos$ to a co-function $\sin$, equate arguments (mod $2\pi$ and the supplementary branch), then square to land on $\sin2\theta=\pm\frac34$.

   Concept and Intuition

   $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$ mixes sine and cosine of different bounded arguments. Rewriting the RHS as a sine via $\cos\phi=\sin(\frac\pi2-\phi)$ lets us use the standard $\sin A=\sin B \Rightarrow A=B$ or $A=\pi-B$ (mod $2\pi$) equivalence, and because $\pi\cos\theta,\pi\sin\theta\in[-\pi,\pi]$ only the $n=0$ branch is achievable.

   Step-by-Step Solution

   1. $\cos(\pi\sin\theta) = \sin\left(\frac{\pi}{2} - \pi\sin\theta\right)$.
   2. So $\sin(\pi\cos\theta) = \sin\left(\frac{\pi}{2}-\pi\sin\theta\right)$.
   3. Branch 1: $\pi\cos\theta = \frac{\pi}{2}-\pi\sin\theta + 2n\pi \Rightarrow \cos\theta+\sin\theta = \frac12+2n$. Since $\cos\theta+\sin\theta\in[-\sqrt2,\sqrt2]$, only $n=0$ works: $\cos\theta+\sin\theta=\frac12$.
   4. Branch 2: $\pi\cos\theta = \pi-\left(\frac{\pi}{2}-\pi\sin\theta\right)+2n\pi \Rightarrow \cos\theta-\sin\theta = \frac12+2n$, and again only $n=0$: $\cos\theta-\sin\theta=\frac12$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.$\tan 81^\circ - \tan 63^\circ - \tan 27^\circ + \tan 9^\circ =$ (A) $2$ (B) $3$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   A telescoping identity using $\cot\theta-\tan\theta=2\cot2\theta$ turns this sum of four tangents of special angles into a clean constant: $4$.

   Concept and Intuition

   Whenever you see $\tan(90^\circ-\theta)-\tan\theta$ patterns, rewrite $\tan(90^\circ-\theta)=\cot\theta$ and use the standard identity $\cot\theta-\tan\theta=\dfrac{\cos^2\theta-\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{2\cos2\theta}{\sin2\theta}=2\cot2\theta$.

   Step-by-Step Solution

   1. $\tan81^\circ=\cot9^\circ$ and $\tan63^\circ=\cot27^\circ$.
   2. So the expression $=(\cot9^\circ+\tan9^\circ)-(\cot27^\circ+\tan27^\circ)$.
   3. Use $\cot\theta+\tan\theta=\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}=\dfrac{2}{\sin2\theta}$.
   4. So expression $=\dfrac{2}{\sin18^\circ}-\dfrac{2}{\sin54^\circ}$. …

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4. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.For $n \in \mathbb{Z}$, a set of values of $\theta$ satisfying $\sec\theta - 1 = (\sqrt{2} - 1)\tan\theta$ is (A) $2n\pi - \dfrac{\pi}{4}$ (B) $2n\pi + \dfrac{\pi}{2}$ (C) $(2n+1)\pi + \dfrac{\pi}{4}$ (D) $2n\pi + \dfrac{\pi}{4}$

   ›Reveal solutionSolution

   Clearing denominators and converting to half-angle form reduces the equation to $\tan(\theta/2)=\tan(\pi/8)$, giving $\theta=2n\pi+\pi/4$.

   Concept and Intuition

   $\sec\theta-1$ and $\tan\theta$ both vanish at $\theta=0$, so factoring through half-angle identities ($1-\cos\theta=2\sin^2(\theta/2)$, $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$) turns a mixed-function equation into a pure $\tan(\theta/2)$ equation.

   Step-by-Step Solution

   1. Multiply both sides by $\cos\theta$: $1-\cos\theta=(\sqrt2-1)\sin\theta$.
   2. Substitute half-angle identities: $2\sin^2\dfrac{\theta}{2}=(\sqrt2-1)\cdot2\sin\dfrac{\theta}{2}\cos\dfrac{\theta}{2}$.
   3. Assuming $\sin(\theta/2)\ne0$ (the trivial branch $\theta=2n\pi$ is not among the answer choices, so we take the genuine oscillating branch): divide through to get $\tan\dfrac{\theta}{2}=\sqrt2-1$.
   4. Recall the exact value $\tan(\pi/8)=\tan22.5^\circ=\sqrt2-1$. …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $u = \log\tan\left(\dfrac{\pi}{4} + \dfrac{\theta}{2}\right)$, then $\tanh\dfrac{u}{2} =$ (A) $\tan\dfrac{\theta}{2}$ (B) $\cot\dfrac{\theta}{2}$ (C) $\sec\dfrac{\theta}{2}$ (D) $\sin\dfrac{\theta}{2}$

   ›Reveal solutionSolution

   This is the classical Gudermannian-function identity: $\tanh(u/2)=\tan(\theta/2)$ when $u=\log\tan(\pi/4+\theta/2)$.

   Concept and Intuition

   The substitution $u=\log\tan(\pi/4+\theta/2)$ links circular and hyperbolic functions (used e.g. in the Mercator map projection). Converting $\tanh(u/2)$ into exponentials of $u$, then substituting $e^u$, collapses everything back to a clean circular function of $\theta/2$.

   Step-by-Step Solution

   1. $\tanh\dfrac{u}{2}=\dfrac{e^{u}-1}{e^{u}+1}$ (standard identity, since $\tanh x=\dfrac{e^{2x}-1}{e^{2x}+1}$ with $x=u/2$).
   2. Given $u=\log\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)$, we have $e^u=\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)$.
   3. Let $t=\tan(\theta/2)$. Then $\tan\left(\dfrac{\pi}{4}+\dfrac{\theta}{2}\right)=\dfrac{1+t}{1-t}$. …

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6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The quadratic equation whose roots are $\cos 72^\circ$ and $\sin 54^\circ$ is (A) $4x^2 + \sqrt{5}x - 1 = 0$ (B) $4x^2 + 2\sqrt{5}x + 1 = 0$ (C) $4x^2 - 2\sqrt{5}x + 1 = 0$ (D) $x^2 - 2\sqrt{5}x + 4 = 0$

   ›Reveal solutionSolution

   This uses the exact surd values of $\cos36^\circ$ and $\cos72^\circ$ (from the golden-ratio pentagon geometry) to build a quadratic from sum and product of roots. The answer is (C).

   Concept and Intuition

   $36^\circ$ and $72^\circ$ are special angles tied to the regular pentagon, and their cosines involve $\sqrt5$ (the golden ratio). Once we recognize $\sin54^\circ=\cos36^\circ$ (co-function identity), both roots become known surds, and building the quadratic is just $x^2-(\text{sum})x+(\text{product})=0$.

   Step-by-Step Solution

   1. $\sin54^\circ = \sin(90^\circ-36^\circ)=\cos36^\circ$.
   2. Known exact values: $\cos36^\circ = \dfrac{\sqrt5+1}{4}\cdot 1$ — precisely $\cos36^\circ = \dfrac{1+\sqrt5}{4}$ (numerically $\approx0.809$, correct), and $\cos72^\circ = \dfrac{\sqrt5-1}{4}$ (numerically $\approx0.309$, correct).
   3. Roots: $r_1=\cos72^\circ = \dfrac{\sqrt5-1}{4}$, $r_2=\sin54^\circ=\cos36^\circ=\dfrac{\sqrt5+1}{4}$.
   4. Sum $= r_1+r_2 = \dfrac{(\sqrt5-1)+(\sqrt5+1)}{4} = \dfrac{2\sqrt5}{4}=\dfrac{\sqrt5}{2}$. …

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7. 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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8. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $A + B + C = \pi$, then $3 - 2\left(\cos\dfrac{A}{2}\cos\dfrac{B}{2}\sin\dfrac{C}{2} + \cos\dfrac{A}{2}\sin\dfrac{B}{2}\cos\dfrac{C}{2} + \sin\dfrac{A}{2}\cos\dfrac{B}{2}\cos\dfrac{C}{2}\right) =$ (A) $\sin^2 A + \sin^2 B + \sin^2 C$ (B) $\cos^2 A + \cos^2 B + \cos^2 C$ (C) $\sin^2\dfrac{A}{2} + \sin^2\dfrac{B}{2} + \sin^2\dfrac{C}{2}$ (D) $\cos^2\dfrac{A}{2} + \cos^2\dfrac{B}{2} + \cos^2\dfrac{C}{2}$

   ›Reveal solutionSolution

   This is a standard triangle trig identity built from the sine-of-sum expansion at half-angles. The answer is (C).

   Concept and Intuition

   When $A+B+C=\pi$, the half-angles satisfy $\frac A2+\frac B2+\frac C2 = \frac\pi2$. Expanding $\sin$ of that sum using the standard three-term product expansion links the bracketed expression directly to $\sin\frac A2\sin\frac B2\sin\frac C2$, and from there to the well-known triangle identity for the sum of $\sin^2$ of the half-angles.

   Step-by-Step Solution

   1. Let $a=\frac A2,\,b=\frac B2,\,c=\frac C2$, so $a+b+c=\frac\pi2$ and $\sin(a+b+c)=1$.
   2. Expand: $\sin(a+b+c) = \sin a\cos b\cos c + \cos a\sin b\cos c + \cos a\cos b\sin c - \sin a\sin b\sin c$.
   3. The given bracket $S = \cos a\cos b\sin c+\cos a\sin b\cos c+\sin a\cos b\cos c$ is exactly the first three terms of this expansion (just reordered). So $1 = S - \sin a\sin b\sin c$, giving $S = 1+\sin a\sin b\sin c$.
   4. The expression asked for is $3-2S = 3-2(1+\sin a\sin b\sin c) = 1-2\sin a\sin b\sin c = 1-2\sin\frac A2\sin\frac B2\sin\frac C2$.
   5. Recall the standard identity: $\sin^2\frac A2+\sin^2\frac B2+\sin^2\frac C2 = 1-2\sin\frac A2\sin\frac B2\sin\frac C2$ for a triangle.
   6. So $3-2S = \sin^2\frac A2+\sin^2\frac B2+\sin^2\frac C2$, matching option (C). …

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9. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $6\sin^2 x = 3\cos^4 x - \sin^2 x \cos^2 x$, then $x =$ (A) $2n\pi \pm \dfrac{\pi}{3}\ \forall n \in \mathbb{Z}$ (B) $n\pi \pm \dfrac{\pi}{3}\ \forall n \in \mathbb{Z}$ (C) $n\pi \pm \dfrac{\pi}{6}\ \forall n \in \mathbb{Z}$ (D) $2n\pi \pm \dfrac{\pi}{4}\ \forall n \in \mathbb{Z}$

   ›Reveal solutionSolution

   Substituting $\cos^2x=1-\sin^2x$ turns the equation into a quadratic in $\sin^2x$, and the valid root gives a standard $\sin^2x=\sin^2\alpha$ solution. The answer is (C).

   Concept and Intuition

   Any equation that's homogeneous in $\sin x,\cos x$ (or reducible via $\cos^2x=1-\sin^2x$) to a polynomial in one trig function can be solved as an algebraic equation first, then converted to a general angle solution. Here, once we know $\sin^2x$ equals a specific value, we use the identity that $\sin^2x=\sin^2\alpha \iff x=n\pi\pm\alpha$.

   Step-by-Step Solution

   1. Let $s=\sin^2x$, so $\cos^2x=1-s$, and $\cos^4x=(1-s)^2$.
   2. The equation $6\sin^2x = 3\cos^4x-\sin^2x\cos^2x$ becomes $6s = 3(1-s)^2 - s(1-s)$.
   3. Expand: $3(1-s)^2 = 3-6s+3s^2$, and $s(1-s)=s-s^2$. So RHS $=3-6s+3s^2-s+s^2 = 3-7s+4s^2$.
   4. So $6s = 3-7s+4s^2 \Rightarrow 4s^2-13s+3=0$.
   5. Solve: $s = \dfrac{13\pm\sqrt{169-48}}{8} = \dfrac{13\pm11}{8}$, giving $s=3$ or $s=\dfrac14$.
   6. Since $s=\sin^2x\le1$, reject $s=3$. So $\sin^2x=\dfrac14 = \sin^2\dfrac\pi6$. …

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10. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.$\dfrac{\cos\theta}{1-\cos\theta} + \dfrac{\sec\theta}{1+\sec\theta} =$ (A) $1 + 2\tan^2\theta$ (B) $\sec^2\theta + \csc^2\theta$ (C) $\tan^2\theta + \cot^2\theta$ (D) $1 + 2\cot^2\theta$

    ›Reveal solutionSolution

    Simplify the second term to a form with the same $(1\pm\cos\theta)$ denominators as the first, combine, and reduce using $\sin^2\theta+\cos^2\theta=1$ and $\csc^2\theta=1+\cot^2\theta$. The sum simplifies to $1+2\cot^2\theta$.

    Concept and Intuition

    Many trig-identity problems become tractable once every term is rewritten in terms of $\sin\theta$ and $\cos\theta$ only — here, $\sec\theta=\frac1{\cos\theta}$ simplifies the second fraction dramatically, revealing a common structure with the first term (both end up over denominators built from $1\pm\cos\theta$, whose product is $\sin^2\theta$).

    Step-by-Step Solution

    1. Simplify the second term: $\dfrac{\sec\theta}{1+\sec\theta}=\dfrac{1/\cos\theta}{1+1/\cos\theta}=\dfrac{1/\cos\theta}{(\cos\theta+1)/\cos\theta}=\dfrac{1}{1+\cos\theta}$.
    2. The expression becomes $\dfrac{\cos\theta}{1-\cos\theta}+\dfrac{1}{1+\cos\theta}$.
    3. Combine over the common denominator $(1-\cos\theta)(1+\cos\theta)=1-\cos^2\theta=\sin^2\theta$: $\frac{\cos\theta(1+\cos\theta) + (1-\cos\theta)}{\sin^2\theta} = \frac{\cos\theta+\cos^2\theta+1-\cos\theta}{\sin^2\theta}=\frac{1+\cos^2\theta}{\sin^2\theta}.$ …

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11. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.For $\theta \in \left(0, \dfrac{\pi}{2}\right)$, if the complete range of $(\cot^2\theta - \cos^2\theta)(\tan^2\theta - \sin^2\theta)$ is $(\alpha, \beta]$ then $\beta - \alpha =$ (A) 1 (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{4}$ (D) 2

    ›Reveal solutionSolution

    Both factors collapse neatly to give the product $\cos^2\theta\sin^2\theta = \frac14\sin^2(2\theta)$, whose range on $(0,\pi/2)$ is $(0,\tfrac14]$ — so $\beta-\alpha=\tfrac14$.

    Concept and Intuition

    Rather than expanding the product directly, it pays to simplify each bracket separately using $\cot\theta=\cos\theta/\sin\theta$ and $\tan\theta=\sin\theta/\cos\theta$ — each bracket turns out to be a perfect "difference into a single power" simplification, and multiplying the two results collapses everything into the well-known double-angle expression $\sin^2(2\theta)/4$, whose range over a given domain is easy to reason about directly (rather than doing calculus on the original messy expression).

    Step-by-Step Solution

    1. Simplify the first bracket:

    $\cot^2\theta-\cos^2\theta = \cos^2\theta\left(\frac{1}{\sin^2\theta}-1\right) = \cos^2\theta\cdot\frac{1-\sin^2\theta}{\sin^2\theta} = \cos^2\theta\cdot\frac{\cos^2\theta}{\sin^2\theta}=\frac{\cos^4\theta}{\sin^2\theta}.$

    2. Simplify the second bracket similarly:

    $\tan^2\theta-\sin^2\theta = \sin^2\theta\left(\frac{1}{\cos^2\theta}-1\right) = \sin^2\theta\cdot\frac{\sin^2\theta}{\cos^2\theta}=\frac{\sin^4\theta}{\cos^2\theta}.$

    3. Multiply the two results: $\frac{\cos^4\theta}{\sin^2\theta}\cdot\frac{\sin^4\theta}{\cos^2\theta} = \cos^2\theta\sin^2\theta = \frac{1}{4}\sin^2(2\theta).$ …

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12. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If $\tan x + \cot x = 6$, then $\tan^3 x + \cot^3 x =$ (A) $192$ (B) $180$ (C) $198$ (D) $186$

    ›Reveal solutionSolution

    Using the algebraic identity for the sum of cubes in terms of the sum, $\tan^3x+\cot^3x = (\tan x+\cot x)^3 - 3(\tan x+\cot x)$, since $\tan x\cdot\cot x=1$ always.

    Concept and Intuition

    Whenever a problem gives you $t+\frac1t$ and asks for $t^3+\frac{1}{t^3}$, the clean route is the algebraic identity $a^3+b^3=(a+b)^3-3ab(a+b)$, applied with $a=t,\ b=1/t$. Because $t\cdot\frac1t=1$ identically, the identity simplifies beautifully to $t^3+\frac{1}{t^3}=(t+\tfrac1t)^3-3(t+\tfrac1t)$ — no need to ever find $\tan x$ itself.

    Step-by-Step Solution

    1. Let $t=\tan x$. Since $\cot x = 1/\tan x = 1/t$, the given condition is $t+\dfrac1t=6$.
    2. We want $t^3+\dfrac{1}{t^3}$. Use $a^3+b^3=(a+b)^3-3ab(a+b)$ with $a=t,\,b=\dfrac1t$, noting $ab=1$. …

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