Q.$\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ is equal to (A) $\tan x + \cot x + C$ (B) $\tan x + \operatorname{cosec} x + C$ (C) $-\tan x + \cot x + C$ (D) $\tan x + \sec x + C$

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-2022-07-06-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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3. 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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4. 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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5. 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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6. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\cos\theta(\csc\theta - \sec\theta) - \cot\theta =$ (A) -1 (B) 1 (C) 0 (D) $\cos^2\theta - \tan^2\theta$

   ›Reveal solutionSolution

   Direct expansion into sines and cosines shows the $\cot\theta$ terms cancel, leaving $-1$.

   Concept and Intuition

   When an expression mixes several different trig ratios ($\csc,\sec,\cot$), it's usually fastest to

   rewrite everything in terms of $\sin\theta$ and $\cos\theta$ and simplify directly, rather than

   hunting for a named identity.

   Step-by-Step Solution

   1. Expand: $\cos\theta(\csc\theta-\sec\theta) = \cos\theta\csc\theta - \cos\theta\sec\theta$.
   2. $\cos\theta\csc\theta = \cos\theta\cdot\dfrac{1}{\sin\theta} = \dfrac{\cos\theta}{\sin\theta}=\cot\theta$.
   3. $\cos\theta\sec\theta = \cos\theta\cdot\dfrac1{\cos\theta}=1$.
   4. So the expression becomes $\cot\theta - 1 - \cot\theta = -1$.

   Common Mistakes …

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7. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.$\dfrac{\tan A}{1-\cot A} + \dfrac{\cot A}{1-\tan A} =$ (A) $\sec A \csc A - 1$ (B) $\tan A + \cot A$ (C) $\tan A + \cot A + 1$ (D) $\sec A + \csc A + 1$

   ›Reveal solutionSolution

   Substituting $t=\tan A$ turns the expression into an algebraic fraction that factors via $t^3-1$, simplifying cleanly to $\tan A+\cot A+1$.

   Concept and Intuition

   Trig identities involving $\tan$ and $\cot$ of the same angle often simplify beautifully after substituting $t=\tan A$ (so $\cot A = 1/t$), turning the problem into ordinary algebra, and using the factorization $t^3-1=(t-1)(t^2+t+1)$.

   Step-by-Step Solution

   1. Let $t=\tan A$, so $\cot A=\dfrac1t$.
   2. First term: $\dfrac{\tan A}{1-\cot A}=\dfrac{t}{1-\frac1t}=\dfrac{t}{\frac{t-1}{t}}=\dfrac{t^2}{t-1}$.
   3. Second term: $\dfrac{\cot A}{1-\tan A}=\dfrac{1/t}{1-t}=\dfrac{1}{t(1-t)}=\dfrac{-1}{t(t-1)}$.
   4. Sum $=\dfrac{t^2}{t-1}-\dfrac{1}{t(t-1)}=\dfrac{t^3-1}{t(t-1)}$. …

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8. 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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9. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.$\sec^2 x + 5\tan x + 5 =$ (A) $(\tan x+2)(\tan x+3)$ (B) $(\tan x+1)(\tan x+5)$ (C) $(\tan x-2)(\tan x-3)$ (D) $(\sin x+2)(\sin x+5)$

   ›Reveal solutionSolution

   Replacing $\sec^2x$ with $1+\tan^2x$ turns the expression into a simple quadratic in $\tan x$ that factors as $(\tan x+2)(\tan x+3)$.

   Concept and Intuition

   The identity $\sec^2x=1+\tan^2x$ is the go-to substitution whenever an expression mixes $\sec^2x$ with linear/quadratic terms in $\tan x$ — it converts the whole thing into an ordinary polynomial factoring problem.

   Step-by-Step Solution

   1. $\sec^2x+5\tan x+5 = (1+\tan^2x)+5\tan x+5 = \tan^2x+5\tan x+6$.
   2. Factor the quadratic $t^2+5t+6$ (with $t=\tan x$): need two numbers multiplying to 6 and summing to 5 — that's 2 and 3.
   3. $t^2+5t+6 = (t+2)(t+3)$, i.e. $(\tan x+2)(\tan x+3)$.

   Common Mistakes …

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10. 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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11. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.$\dfrac{1+\cos\theta-\sin\theta}{1+\cos\theta+\sin\theta} + \dfrac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta} =$ (A) $2\sec\theta$ (B) $2\,\mathrm{cosec}\,\theta$ (C) $2\tan\theta$ (D) $2\cot\theta$

    ›Reveal solutionSolution

    Writing the sum of the two reciprocal fractions as $\frac{2(a^2+b^2)}{a^2-b^2}$ with $a=1+\cos\theta,\,b=\sin\theta$ collapses neatly to $2\sec\theta$.

    Concept and Intuition

    Whenever an expression has the form $\frac{a-b}{a+b}+\frac{a+b}{a-b}$, combining over a common denominator immediately gives $\frac{(a-b)^2+(a+b)^2}{(a+b)(a-b)}=\frac{2(a^2+b^2)}{a^2-b^2}$ — this avoids messy direct combination and lets the trig identities do the work only once at the end.

    Step-by-Step Solution

    1. Let $a=1+\cos\theta$, $b=\sin\theta$. The expression is $\dfrac{a-b}{a+b}+\dfrac{a+b}{a-b}=\dfrac{(a-b)^2+(a+b)^2}{(a+b)(a-b)}=\dfrac{2a^2+2b^2}{a^2-b^2}$.
    2. Numerator: $2\left[(1+\cos\theta)^2+\sin^2\theta\right]=2\left[1+2\cos\theta+\cos^2\theta+\sin^2\theta\right]=2[2+2\cos\theta]=4(1+\cos\theta)$. …

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12. 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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