Q.Evaluate: $\int \dfrac{\sin^6 x+\cos^6 x}{\sin^2 x\,\cos^2 x}\,dx$

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1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.The value of the product $\cot 10^\circ \cot 20^\circ \cot 30^\circ \cot 45^\circ \cot 60^\circ \cot 70^\circ \cot 80^\circ$ is equal to (A) $\sqrt{3}$ (B) $\dfrac{\sqrt{3}}{3}$ (C) 3 (D) $3\sqrt{3}$ (E) 1

   ›Reveal solutionSolution

   Each complementary pair multiplies to $1$ (since $\cot(90^\circ-\theta)=\tan\theta$), and $\cot45^\circ=1$, giving product $1$.

   Use $\cot(90^\circ-\theta)=\tan\theta$, so $\cot\theta\cot(90^\circ-\theta)=\cot\theta\tan\theta=1$. Pairing: …

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2. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If $\tan\alpha=\dfrac{1}{2}$, then the value of $\tan^2(2\alpha)\sec^2(2\alpha)$ is equal to (A) $\dfrac{200}{9}$ (B) $\dfrac{400}{9}$ (C) $\dfrac{200}{81}$ (D) $\dfrac{400}{81}$ (E) $\dfrac{200}{27}$

   ›Reveal solutionSolution

   Find $\tan2\alpha=\tfrac43$; then $\tan^2 2\alpha\,\sec^2 2\alpha=\tfrac{16}{9}\cdot\tfrac{25}{9}=\tfrac{400}{81}$.

   With $\tan\alpha=\tfrac12$, the double-angle formula gives

   $\tan2\alpha=\frac{2\tan\alpha}{1-\tan^2\alpha}=\frac{2\cdot\frac12}{1-\frac14}=\frac{1}{3/4}=\frac43.$ …

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3. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If $(3\cos x - 2\sec x)^2 = 9\cos^2 x + 4\tan^2 x + k$ , where $k$ is a constant, then the value of $k$ is equal to (A) 12 (B) -12 (C) -4 (D) 8 (E) -8

   ›Reveal solutionSolution

   Expanding and using $\sec^2x=1+\tan^2x$ turns the middle term into $-12$ plus $4+4\tan^2x$, leaving $k=-8$.

   Expand:

   $(3\cos x-2\sec x)^2=9\cos^2x-2(3\cos x)(2\sec x)+4\sec^2x=9\cos^2x-12\cos x\sec x+4\sec^2x.$

   Since $\cos x\sec x=1$, the middle term is $-12$. Using $\sec^2x=1+\tan^2x$, $4\sec^2x=4+4\tan^2x$. Thus …

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4. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If $\alpha$ and $\beta$ are real constants such that $\alpha - \beta = \dfrac{\pi}{4}$ , then the value of $(\sin\alpha + \sin\beta)^2 + (\cos\alpha + \cos\beta)^2$ is equal to (A) $2$ (B) $\sqrt{2}$ (C) $2\sqrt{2}$ (D) $2+\sqrt{2}$ (E) $2-\sqrt{2}$

   ›Reveal solutionSolution

   The expression equals $2+2\cos(\alpha-\beta)=2+2\cos\tfrac{\pi}{4}=2+\sqrt2$.

   Expand both squares:

   $(\sin\alpha+\sin\beta)^2+(\cos\alpha+\cos\beta)^2=(\sin^2\alpha+\cos^2\alpha)+(\sin^2\beta+\cos^2\beta)+2(\sin\alpha\sin\beta+\cos\alpha\cos\beta).$

   This is

   $1+1+2\cos(\alpha-\beta)=2+2\cos(\alpha-\beta).$ …

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5. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If $\csc t+\cot t=\dfrac{5}{2}$, then the value of $\tan t$ is equal to (A) $\dfrac{10}{21}$ (B) $\dfrac{21}{20}$ (C) $\dfrac{20}{21}$ (D) $\dfrac{21}{10}$ (E) $\dfrac{10}{42}$

   ›Reveal solutionSolution

   The identity gives $\csc t-\cot t=\tfrac25$; solving, $\cot t=\tfrac{21}{20}$ and $\tan t=\tfrac{20}{21}$.

   Since $(\csc t+\cot t)(\csc t-\cot t)=\csc^{2}t-\cot^{2}t=1$ and $\csc t+\cot t=\tfrac52$,

   $\csc t-\cot t=\frac{1}{5/2}=\frac25.$

   Subtracting the two equations: …

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6. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.$\sin 15^{\circ}\sin 45^{\circ}\sin 75^{\circ} =$ (A) $\dfrac{\sqrt{2}}{4}$ (B) $\dfrac{\sqrt{2}}{2}$ (C) $\dfrac{\sqrt{2}}{8}$ (D) $-\dfrac{\sqrt{2}}{8}$ (E) $-\dfrac{\sqrt{2}}{4}$

   ›Reveal solutionSolution

   Pair $15^\circ$ and $75^\circ$: their product is $\tfrac14$; multiply by $\sin45^\circ$ to get $\tfrac{\sqrt2}{8}$.

   Since $\sin75^\circ=\cos15^\circ$,

   $\sin15^\circ\sin75^\circ=\sin15^\circ\cos15^\circ=\tfrac12\sin30^\circ=\tfrac12\cdot\tfrac12=\tfrac14.$

   Then …

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7. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\sin\left(x+\dfrac{7\pi}{4}\right)+\sin\left(x-\dfrac{7\pi}{4}\right)$ is equal to (A) $\sqrt{2}\sin x$ (B) $-\sqrt{2}\cos x$ (C) $-\sqrt{2}\sin x$ (D) $\sqrt{2}\cos x$ (E) $-2\cos x$

   ›Reveal solutionSolution

   Sum-to-product gives $2\sin x\cos\tfrac{7\pi}{4}=\sqrt2\sin x$.

   Using $\sin(A+B)+\sin(A-B)=2\sin A\cos B$ with $A=x$, $B=\tfrac{7\pi}{4}$:

   $2\sin x\cos\tfrac{7\pi}{4}.$ …

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8. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.Let $f(x) = 2\sqrt{2}\sin x + 2\sqrt{2}\cos x,\ x \in \mathbb{R}$. Then the value of $f\left(\dfrac{\pi}{12}\right)$ is equal to (A) $\sqrt{3}$ (B) $2\sqrt{3}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\sqrt{2}$ (E) $2\sqrt{2}$

   ›Reveal solutionSolution

   Write $f(x)=4\sin(x+\tfrac\pi4)$; at $x=\tfrac\pi{12}$ this is $4\sin\tfrac\pi3=2\sqrt3$.

   $f(x)=2\sqrt2\sin x+2\sqrt2\cos x=2\sqrt2\cdot\sqrt2\,\sin\!\big(x+\tfrac{\pi}{4}\big)=4\sin\!\big(x+\tfrac{\pi}{4}\big).$

   At $x=\tfrac{\pi}{12}$: …

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9. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The value of $sin6^{o}cos36^{o}sin66^{o} + cos12^{o}sin42^{o}sin18^{o}$ is equal to (A) $\frac{1}{12}(sin18^{o} + cos36^{o})$ (B) $\frac{1}{3}(sin18^{o} + cos36^{o})$ (C) $\frac{1}{16}(sin18^{o} + cos36^{o})$ (D) $\frac{1}{4}(sin18^{o} + cos36^{o})$ (E) $\frac{1}{2}(sin18^{o} + cos36^{o})$

   ›Reveal solutionSolution

   Evaluate the products; the sum is exactly $\tfrac14(\sin18^\circ+\cos36^\circ)$.

   Compute each term:

   $\sin6^\circ\cos36^\circ\sin66^\circ = 0.10453\times0.80902\times0.91355 = 0.07726$.

   $\cos12^\circ\sin42^\circ\sin18^\circ = 0.97815\times0.66913\times0.30902 = 0.20226$.

   Sum $=0.27951$. …

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10. KEAM 2026Set eng-2026-04194 marksMCQ
    Q.The value of $sin\left(2sin^{-1}\frac{3}{5}\right)$ is equal to (A) $\frac{23}{25}$ (B) $\frac{21}{25}$ (C) $\frac{22}{25}$ (D) $\frac{24}{25}$ (E) $\frac{18}{25}$

    ›Reveal solutionSolution

    Set $\theta=\sin^{-1}\tfrac35$; then $\sin(2\theta)=2\sin\theta\cos\theta=\tfrac{24}{25}$.

    Let $\theta=\sin^{-1}\tfrac35$, so $\sin\theta=\tfrac35$ and $\cos\theta=\sqrt{1-\tfrac{9}{25}}=\tfrac45$. …

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11. KEAM 2026Set eng-2026-04204 marksMCQ
    Q.$\sin\left(\dfrac{3\pi}{4} + x\right) + \cos\left(\dfrac{3\pi}{4} + x\right) - \sin\left(\dfrac{3\pi}{4} - x\right) - \cos\left(\dfrac{3\pi}{4} - x\right) =$ (A) $-2\sqrt{2}\sin x$ (B) $\sqrt{2}\sin x$ (C) $2\sqrt{2}\sin x$ (D) $3\sqrt{2}\sin x$ (E) $-\sqrt{2}\sin x$

    ›Reveal solutionSolution

    Group the sine terms and cosine terms and apply the difference-to-product identities. Each pair evaluates to $-\sqrt2\sin x$ (since $\cos\frac{3\pi}{4}=-\frac{1}{\sqrt2}$ and $\sin\frac{3\pi}{4}=\frac{1}{\sqrt2}$), summing to $-2\sqrt2\sin x$.

    Group as $[\sin(\tfrac{3\pi}{4}+x)-\sin(\tfrac{3\pi}{4}-x)]+[\cos(\tfrac{3\pi}{4}+x)-\cos(\tfrac{3\pi}{4}-x)]$.

    Using $\sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2}$:

    $\sin(\tfrac{3\pi}{4}+x)-\sin(\tfrac{3\pi}{4}-x)=2\cos\tfrac{3\pi}{4}\sin x=2\left(-\tfrac{1}{\sqrt2}\right)\sin x=-\sqrt2\sin x.$ …

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12. KEAM 2026Set eng-2026-04204 marksMCQ
    Q.$2^{11}\sin(\frac{x}{2^{10}})\cos(\frac{x}{2})\cos(\frac{x}{2^2})\cos(\frac{x}{2^3})\ldots\cos(\frac{x}{2^{10}})$ is equal to (A) $\sin x$ (B) $2\sin x$ (C) $-2\sin x$ (D) $4\sin x$ (E) $8\sin x$

    ›Reveal solutionSolution

    Repeated use of $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$ collapses the cosine product to $\frac{\sin x}{2^{10}\sin(x/2^{10})}$. The prefactor $2^{11}\sin(x/2^{10})$ then leaves $2\sin x$.

    The telescoping identity for a product of cosines is

    $\prod_{k=1}^{n}\cos\frac{x}{2^{k}}=\frac{\sin x}{2^{n}\sin\left(x/2^{n}\right)}.$

    Here the product runs $k=1$ to $10$:

    $\cos\frac{x}{2}\cos\frac{x}{2^2}\cdots\cos\frac{x}{2^{10}}=\frac{\sin x}{2^{10}\sin\left(x/2^{10}\right)}.$ …

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