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

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.$1 + \sec^2 x \sin^2 x =$ (A) $\sin 2x$ (B) $\sin^2 x$ (C) $\tan^2 x$ (D) $\sec^2 x$

   ›Reveal solutionSolution

   Rewrite $\sec^2x\sin^2x$ as $\tan^2x$, then apply the standard identity $1+\tan^2x=\sec^2x$.

   Concept and Intuition

   This is a direct application of the fundamental Pythagorean trigonometric identities. Recognising $\sec^2x\sin^2x$ as $\tan^2x$ (since $\sec x = 1/\cos x$) is the key simplification step.

   Step-by-Step Solution

   1. $\sec^2x\sin^2x = \dfrac{1}{\cos^2x}\cdot\sin^2x = \dfrac{\sin^2x}{\cos^2x} = \tan^2x$.
   2. So the expression becomes $1+\tan^2x$.
   3. By the Pythagorean identity, $1+\tan^2x = \sec^2x$.

   Common Mistakes …

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2. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.$\tan x + \dfrac{\cos x}{1+\sin x} =$ (A) $\tan 2x$ (B) $\operatorname{cosec} x$ (C) $\sec x$ (D) $\cos 2x$

   ›Reveal solutionSolution

   Combine the two terms over a common denominator and use $\sin^2x+\cos^2x=1$ to collapse everything to $\sec x$.

   Concept and Intuition

   Whenever you see a sum of a trig ratio and a fraction like $\dfrac{\cos x}{1+\sin x}$, the standard move is to put both terms over the common denominator $\cos x(1+\sin x)$ and simplify the numerator using the Pythagorean identity.

   Step-by-Step Solution

   1. Write $\tan x = \dfrac{\sin x}{\cos x}$.
   2. Common denominator: $\dfrac{\sin x(1+\sin x) + \cos x\cdot\cos x}{\cos x(1+\sin x)} = \dfrac{\sin x + \sin^2 x + \cos^2 x}{\cos x(1+\sin x)}$.
   3. Since $\sin^2x+\cos^2x=1$: numerator $= \sin x + 1$. …

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

   ›Reveal solutionSolution

   Rationalizing the fraction using $1-\sin^2x=\cos^2x$ turns it into $\sec x - \tan x$, which cancels the added $\tan x$ to leave $\sec x$.

   Concept and Intuition

   Expressions with $1+\sin x$ in the denominator are cleanly handled by multiplying by the conjugate $1-\sin x$, since $(1+\sin x)(1-\sin x)=1-\sin^2x=\cos^2x$ — this is the same trick used to rationalize surds, applied to trig.

   Step-by-Step Solution

   1. Multiply $\dfrac{\cos x}{1+\sin x}$ by $\dfrac{1-\sin x}{1-\sin x}$: numerator becomes $\cos x(1-\sin x)$, denominator becomes $1-\sin^2x=\cos^2x$.
   2. Simplify: $\dfrac{\cos x(1-\sin x)}{\cos^2 x} = \dfrac{1-\sin x}{\cos x} = \dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x} = \sec x - \tan x$. …

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5. 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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6. 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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7. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $2\sin x - \cos 2x = 1$, then $(3-2\sin^2 x) =$ (A) $\sqrt{3}$ (B) $-\sqrt{3}$ (C) $\sqrt{5}$ (D) $-\sqrt{5}$

   ›Reveal solutionSolution

   Rewrite $\cos2x$ in terms of $\sin x$ to get a quadratic in $\sin x$, solve it, and substitute back into the target expression. Answer: $\sqrt5$.

   Concept and Intuition

   Many trig equations reduce to a quadratic once everything is expressed in a single function (here $\sin x$). Once that quadratic relation is known, expressions like $3-2\sin^2x$ can often be evaluated directly from the quadratic itself, without solving for $x$ numerically.

   Step-by-Step Solution

   1. Use $\cos2x=1-2\sin^2x$. The equation $2\sin x-\cos2x=1$ becomes $2\sin x-(1-2\sin^2x)=1$.
   2. Simplify: $2\sin x-1+2\sin^2x=1\Rightarrow2\sin^2x+2\sin x-2=0\Rightarrow\sin^2x+\sin x-1=0$.
   3. From this, $\sin^2x=1-\sin x$ — this relation lets us substitute directly into the target expression.
   4. Target: $3-2\sin^2x=3-2(1-\sin x)=3-2+2\sin x=1+2\sin x$. …

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8. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.In any triangle ABC, $\frac{\cos 2A}{a^2} - \frac{\cos 2B}{b^2} =$ (A) $a^2 - b^2$ (B) $\frac{1}{a^2} - \frac{1}{b^2}$ (C) $a^2 + b^2$ (D) $\frac{1}{a^2} + \frac{1}{b^2}$

   ›Reveal solutionSolution

   Expressing $\cos2A$ via the law of sines turns the expression into $\frac1{a^2}-\frac1{2R^2}$ minus

   its $B$-counterpart, and the circumradius terms cancel, leaving $\dfrac1{a^2}-\dfrac1{b^2}$.

   Concept and Intuition

   Whenever a triangle identity mixes a double angle (like $\cos2A$) with a side length, the law of

   sines ($a=2R\sin A$) is the natural bridge — it converts the angle into the side via the common

   circumradius $R$, which is exactly what's needed here since $R$ ends up cancelling between the two terms.

   Step-by-Step Solution

   1. Law of sines: $a=2R\sin A \Rightarrow \sin A = \dfrac{a}{2R} \Rightarrow \sin^2A=\dfrac{a^2}{4R^2}$.
   2. Double-angle: $\cos2A = 1-2\sin^2A = 1-\dfrac{2a^2}{4R^2} = 1-\dfrac{a^2}{2R^2}$.
   3. So $\dfrac{\cos2A}{a^2} = \dfrac{1}{a^2}-\dfrac{1}{2R^2}$.
   4. Likewise $\dfrac{\cos2B}{b^2} = \dfrac1{b^2}-\dfrac1{2R^2}$.
   5. Subtract: $\dfrac{\cos2A}{a^2}-\dfrac{\cos2B}{b^2} = \left(\dfrac1{a^2}-\dfrac1{2R^2}\right)-\left(\dfrac1{b^2}-\dfrac1{2R^2}\right) = \dfrac1{a^2}-\dfrac1{b^2}$. …

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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-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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11. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The general solution of $2\cos^2 x - 2\tan x + 1 = 0$ is (A) $n\pi + \dfrac{\pi}{4}, n \in Z$ (B) $2n\pi \pm \dfrac{\pi}{4}, n \in Z$ (C) $2n\pi \pm \dfrac{\pi}{3}, n \in Z$ (D) $n\pi \pm \dfrac{\pi}{3}, n \in Z$

    ›Reveal solutionSolution

    Substituting $t=\tan x$ converts the mixed $\cos^2x/\tan x$ equation into a cubic in $t$ with a unique real root $t=1$, giving $x=n\pi+\dfrac{\pi}{4}$.

    Concept and Intuition

    $\cos^2x$ and $\tan x$ can both be written purely in terms of $t=\tan x$ via $\cos^2x=\dfrac{1}{1+\tan^2x}$. This turns a trigonometric equation into an ordinary polynomial equation, which is easier to solve exactly.

    Step-by-Step Solution

    1. Substitute $\cos^2x=\dfrac{1}{1+t^2}$: $\dfrac{2}{1+t^2}-2t+1=0$.
    2. Multiply through by $(1+t^2)$: $2-2t(1+t^2)+(1+t^2)=0 \Rightarrow 2-2t-2t^3+1+t^2=0$.
    3. Rearranged: $-2t^3+t^2-2t+3=0$, i.e. $2t^3-t^2+2t-3=0$.
    4. Test small rational roots: $t=1$ gives $2(1)-1+2-3=0$ — a root.
    5. Divide out $(t-1)$: $2t^3-t^2+2t-3=(t-1)(2t^2+t+3)$.
    6. Discriminant of $2t^2+t+3$ is $1^2-4(2)(3)=1-24=-23<0$, so no further real roots. …

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12. 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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