Q.Find the value of the expression $\sin\left(2\tan^{-1}\frac{1}{3}\right)+\cos\left(\tan^{-1}2\sqrt2\right)$.

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\sin^2\frac{2\pi}{3} + \cos^2\frac{5\pi}{6} - \tan^2\frac{3\pi}{4} =$ (A) $0$ (B) $\frac{1}{2}$ (C) $1$ (D) $\frac{1}{3}$

   ›Reveal solutionSolution

   Evaluating each of the three standard trig values at its reference angle and combining gives $\frac34+\frac34-1=\frac12$.

   Concept and Intuition

   All three angles here ($\frac{2\pi}{3}=120^\circ$, $\frac{5\pi}{6}=150^\circ$, $\frac{3\pi}{4}=135^\circ$) are standard second-quadrant angles whose trig values reduce to familiar $30^\circ$-$45^\circ$-$60^\circ$ reference values (with appropriate signs for QII). Squaring removes any sign ambiguity for the first two terms; only the $\tan^2$ term needs the correct sign of $\tan$ before squaring (though squaring makes it positive too, so it doesn't actually matter here — but it's still good practice to track the sign).

   Step-by-Step Solution

   1. $\sin\dfrac{2\pi}{3} = \sin120^\circ = \sin(180^\circ-60^\circ) = \sin60^\circ = \dfrac{\sqrt3}{2}$, so $\sin^2\dfrac{2\pi}{3} = \dfrac{3}{4}$.
   2. $\cos\dfrac{5\pi}{6} = \cos150^\circ = \cos(180^\circ-30^\circ) = -\cos30^\circ = -\dfrac{\sqrt3}{2}$, so $\cos^2\dfrac{5\pi}{6} = \dfrac{3}{4}$.
   3. $\tan\dfrac{3\pi}{4} = \tan135^\circ = \tan(180^\circ-45^\circ) = -\tan45^\circ = -1$, so $\tan^2\dfrac{3\pi}{4} = 1$. …

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2. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\sin 22\tfrac{1}{2}^\circ =$ (A) $\sqrt{\dfrac{2+\sqrt{2}}{4}}$ (B) $\dfrac{2+\sqrt{2}}{4}$ (C) $\sqrt{\dfrac{2-\sqrt{2}}{4}}$ (D) $\dfrac{2-\sqrt{2}}{4}$

   ›Reveal solutionSolution

   A direct half-angle substitution with $\theta=45^\circ$ gives (C) $\sqrt{\dfrac{2-\sqrt2}{4}}$.

   Concept and Intuition

   $22.5^\circ$ is half of $45^\circ$, a known angle, so the half-angle identity for sine converts the problem into evaluating $\cos45^\circ$, which is standard.

   Step-by-Step Solution

   1. Half-angle identity: $\sin\left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{1-\cos\theta}{2}}$ (positive root since $22.5^\circ$ is in the first quadrant).
   2. Take $\theta = 45^\circ$, so $\theta/2 = 22.5^\circ$.
   3. $\cos45^\circ = \dfrac{\sqrt2}{2}$. …

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3. 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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4. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\dfrac{\sin 1^\circ + \sin 2^\circ + \ldots + \sin 89^\circ}{2(\cos 1^\circ + \cos 2^\circ + \ldots + \cos 44^\circ)+1} =$ (A) $\sqrt{2}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $2$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   Applying the standard sum formulas for $\sum\sin k\theta$ and $\sum\cos k\theta$ (with $\theta=1°$) and simplifying gives the ratio $1/\sqrt2$.

   Concept and Intuition

   Sums of sines/cosines in arithmetic progression of angle have closed forms:

   $\sum_{k=1}^n \sin k\theta = \frac{\sin(n\theta/2)\sin\left(\frac{(n+1)\theta}2\right)}{\sin(\theta/2)},\qquad \sum_{k=1}^n\cos k\theta=\frac{\sin(n\theta/2)\cos\left(\frac{(n+1)\theta}2\right)}{\sin(\theta/2)}$

   These come from telescoping products-to-sums (or multiplying by $2\sin(\theta/2)$ and telescoping).

   Step-by-Step Solution

   1. Numerator: $\sum_{k=1}^{89}\sin k° = \dfrac{\sin(44.5°)\sin(45°)}{\sin(0.5°)}$.
   2. The cosine part of the denominator: $\sum_{k=1}^{44}\cos k° = \dfrac{\sin(22°)\cos(22.5°)}{\sin(0.5°)}$.
   3. So denominator $= 2\cdot\dfrac{\sin22°\cos22.5°}{\sin0.5°}+1$. …

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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 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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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 2024Set eng-2024-05-22-FN1 markMCQ
   Q.In a triangle ABC, if A, B, C are in arithmetic progression and $\cos A + \cos B + \cos C = \dfrac{1+\sqrt{2}+\sqrt{3}}{2\sqrt{2}}$, then $\tan A =$ (A) $\sqrt{3}$ (B) $2 + \sqrt{3}$ (C) $1$ (D) $2 - \sqrt{3}$

   ›Reveal solutionSolution

   The AP condition fixes $B=60^\circ$; the sum-to-product identity for $\cos A+\cos C$ then pins down $A-C=30^\circ$, giving $A=75^\circ$ and $\tan A=2+\sqrt3$.

   Concept and Intuition

   "$A,B,C$ in AP" for triangle angles always forces the middle angle to be $60^\circ$ (since $2B=A+C$ and $A+B+C=180^\circ$ gives $3B=180^\circ$). The given cosine sum can then be handled by isolating $\cos A+\cos C$ and applying the sum-to-product formula, which naturally involves $\cos\frac{A+C}{2}$ — already known since $A+C=120^\circ$ is fixed.

   Step-by-Step Solution

   1. $A,B,C$ in AP $\Rightarrow 2B=A+C$. With $A+B+C=180^\circ$: $3B=180^\circ\Rightarrow B=60^\circ$, and $A+C=120^\circ$.
   2. $\cos A+\cos B+\cos C=\dfrac{1+\sqrt2+\sqrt3}{2\sqrt2}$, and $\cos B=\cos60^\circ=\dfrac12=\dfrac{\sqrt2}{2\sqrt2}$.
   3. So $\cos A+\cos C=\dfrac{1+\sqrt2+\sqrt3}{2\sqrt2}-\dfrac{\sqrt2}{2\sqrt2}=\dfrac{1+\sqrt3}{2\sqrt2}$.
   4. Sum-to-product: $\cos A+\cos C=2\cos\dfrac{A+C}{2}\cos\dfrac{A-C}{2}=2\cos60^\circ\cos\dfrac{A-C}{2}=\cos\dfrac{A-C}{2}$.
   5. So $\cos\dfrac{A-C}{2}=\dfrac{1+\sqrt3}{2\sqrt2}\approx0.9659\Rightarrow\dfrac{A-C}{2}=15^\circ\Rightarrow A-C=30^\circ$. …

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9. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ =$ (A) $\dfrac{3}{4}$ (B) 1 (C) 0 (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   The product $\tan6^\circ\tan42^\circ\tan66^\circ\tan78^\circ$ evaluates exactly to 1.

   Concept and Intuition

   Products of tangents at angles related by $60^\circ$ shifts often collapse via the identity $\tan\theta\tan(60^\circ-\theta)\tan(60^\circ+\theta)=\tan3\theta$. Here, $18^\circ,42^\circ,78^\circ$ fit the pattern $\theta=18^\circ,\,60-\theta=42^\circ,\,60+\theta=78^\circ$, which lets us replace three of the four factors by a single tangent, and the remaining structure resolves to exactly 1 (confirmable numerically to high precision).

   Step-by-Step Solution

   1. Identify the sub-product $\tan42^\circ\tan78^\circ$ as part of the triple $\tan18^\circ\tan42^\circ\tan78^\circ=\tan(3\times18^\circ)=\tan54^\circ$ (using $18,60-18=42,60+18=78$).
   2. So $\tan42^\circ\tan78^\circ = \dfrac{\tan54^\circ}{\tan18^\circ}$.
   3. The full product becomes $\tan6^\circ\tan66^\circ\cdot\dfrac{\tan54^\circ}{\tan18^\circ}$.
   4. Using known values ($\tan54^\circ=\cot36^\circ$, and the relationships among $6^\circ,18^\circ,36^\circ,66^\circ$ following from the standard $18^\circ/36^\circ$ golden-ratio tangent/cotangent identities), this combination simplifies exactly to 1. …

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10. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.The value of $\dfrac{\sin\theta + \sin3\theta}{\cos\theta + \cos3\theta}$ is (A) $\cos2\theta$ (B) $\cot2\theta$ (C) $\tan2\theta$ (D) $\csc\theta + \sin\theta$

    ›Reveal solutionSolution

    Applying the sum-to-product identities to both numerator and denominator, the common factor $\cos\theta$ cancels, leaving $\tan2\theta$.

    Concept and Intuition

    Sum-to-product formulas convert a sum of sines (or cosines) at symmetric angles ($\theta$ and $3\theta$, straddling the midpoint $2\theta$) into a product involving that midpoint angle — a very common simplification trick.

    Step-by-Step Solution

    1. $\sin\theta + \sin3\theta = 2\sin\left(\dfrac{\theta+3\theta}{2}\right)\cos\left(\dfrac{3\theta-\theta}{2}\right) = 2\sin2\theta\cos\theta$.
    2. $\cos\theta + \cos3\theta = 2\cos\left(\dfrac{\theta+3\theta}{2}\right)\cos\left(\dfrac{3\theta-\theta}{2}\right) = 2\cos2\theta\cos\theta$. …

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11. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.What is the value of $\cos\left(22\dfrac{1}{2}\right)^{\circ} =$ (A) $\sqrt{\dfrac{\sqrt{2}-1}{2\sqrt{2}}}$ (B) $\sqrt{\dfrac{\sqrt{2}+1}{2\sqrt{2}}}$ (C) $\sqrt{2} - 1$ (D) $\sqrt{2} + 1$

    ›Reveal solutionSolution

    Applying the half-angle cosine formula to $45^\circ$ and simplifying matches option (B) after rationalizing.

    Concept and Intuition

    $22.5^\circ$ is half of $45^\circ$, a standard angle. The half-angle identity $\cos\left(\dfrac{x}{2}\right)=\sqrt{\dfrac{1+\cos x}{2}}$ (positive root since $22.5^\circ$ is in the first quadrant) converts an unfamiliar angle into known values of $\cos45^\circ$.

    Step-by-Step Solution

    1. Write $22.5^\circ = \dfrac{45^\circ}{2}$.
    2. Apply the half-angle formula: $\cos(22.5^\circ) = \sqrt{\dfrac{1+\cos45^\circ}{2}}$.
    3. Substitute $\cos45^\circ = \dfrac{\sqrt2}{2}$: $\cos(22.5^\circ) = \sqrt{\dfrac{1+\frac{\sqrt2}{2}}{2}} = \sqrt{\dfrac{2+\sqrt2}{4}}$.
    4. Rationalize option (B): $\dfrac{\sqrt2+1}{2\sqrt2}\cdot\dfrac{\sqrt2}{\sqrt2} = \dfrac{2+\sqrt2}{4}$ — identical to step 3. …

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12. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.$\cos^4\dfrac{\pi}{24} - \sin^4\dfrac{\pi}{24} =$ (A) $\dfrac{\sqrt2-\sqrt3}{2}$ (B) $\dfrac{\sqrt2+\sqrt3}{2}$ (C) $\dfrac{\sqrt2-\sqrt6}{4}$ (D) $\dfrac{\sqrt2+\sqrt6}{4}$

    ›Reveal solutionSolution

    Tests the factoring identity $\cos^4\theta-\sin^4\theta=\cos2\theta$; answer reduces to $\cos15^\circ=\dfrac{\sqrt2+\sqrt6}{4}$.

    Concept and Intuition

    A difference of squares $a^4-b^4=(a^2-b^2)(a^2+b^2)$ applies directly here with $a=\cos\theta, b=\sin\theta$; since $\cos^2\theta+\sin^2\theta=1$ always, the whole expression collapses to just $\cos^2\theta-\sin^2\theta=\cos2\theta$ — a huge simplification before plugging in the angle.

    Step-by-Step Solution

    1. $\cos^4\theta-\sin^4\theta=(\cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^2\theta)$.
    2. Since $\cos^2\theta+\sin^2\theta=1$: expression $=\cos^2\theta-\sin^2\theta=\cos2\theta$.
    3. Here $\theta=\pi/24$, so $2\theta=\pi/12=15^\circ$. …

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