Q.Show that $\cos\left(2\tan^{-1}\frac{1}{7}\right)=\sin\left(4\tan^{-1}\frac{1}{3}\right)$.

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $A=\dfrac{\pi}{24}$, then $\dfrac{\cos A+\cos 3A+\cos 5A+\cos 7A}{\sin A+\sin 3A+\sin 5A+\sin 7A}=$ (A) $\sqrt{3}$ (B) $2\sqrt{3}$ (C) $\dfrac{1}{\sqrt{3}}$ (D) $\dfrac{2}{\sqrt{3}}$

   ›Reveal solutionSolution

   Pairing terms symmetric about the middle and applying sum-to-product formulas collapses the whole ratio to $\cot(4A)$, which equals $\cot30^\circ=\sqrt3$ for $A=\pi/24$.

   Concept and Intuition

   When a sum of cosines (or sines) has angles in arithmetic progression, pairing the first-and-last and middle terms via sum-to-product formulas often produces a common factor that cancels between numerator and denominator, collapsing a messy four-term ratio into a single trig function.

   Step-by-Step Solution

   1. Numerator: $\cos A+\cos7A = 2\cos4A\cos3A$ and $\cos3A+\cos5A=2\cos4A\cos A$. Sum: $2\cos4A(\cos3A+\cos A) = 2\cos4A\cdot2\cos2A\cos A = 4\cos4A\cos2A\cos A$.
   2. Denominator: $\sin A+\sin7A=2\sin4A\cos3A$ and $\sin3A+\sin5A=2\sin4A\cos A$. Sum: $2\sin4A(\cos3A+\cos A) = 4\sin4A\cos2A\cos A$.
   3. Ratio: $\dfrac{4\cos4A\cos2A\cos A}{4\sin4A\cos2A\cos A} = \dfrac{\cos4A}{\sin4A} = \cot4A$. …

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2. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If two angles $\alpha, \beta$ are such that $0 < \alpha, \beta < \dfrac{\pi}{4}$, $\sqrt{1+\cos 2\alpha} = \dfrac{3}{\sqrt{5}}$ and $\dfrac{\sqrt{1-\cos 2\beta}}{\sqrt{1+\cos 2\beta}} = \dfrac{1}{7}$, then $(2\alpha+\beta) =$ (A) $\dfrac{\pi}{3}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{3\pi}{4}$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   Extracting $\tan\alpha=1/3$ and $\tan\beta=1/7$ from the given radical expressions, then computing $\tan(2\alpha+\beta)=1$ within the valid angle range, pins $2\alpha+\beta=\pi/4$.

   Concept and Intuition

   $\sqrt{1+\cos2\theta}=\sqrt{2\cos^2\theta}=\sqrt2\,|\cos\theta|$ and $\sqrt{1-\cos2\theta}=\sqrt{2\sin^2\theta}=\sqrt2\,|\sin\theta|$ — these are the standard half-angle-style simplifications. Since both $\alpha,\beta\in(0,\pi/4)$, all trig ratios involved are positive, so absolute values can be dropped safely.

   Step-by-Step Solution

   1. $\sqrt{1+\cos2\alpha}=\sqrt{2\cos^2\alpha}=\sqrt2\cos\alpha$ (positive since $\alpha\in(0,\pi/4)$). Set equal to $\dfrac3{\sqrt5}$: $\cos\alpha=\dfrac3{\sqrt5\sqrt2}=\dfrac3{\sqrt{10}}$.
   2. Then $\sin\alpha=\sqrt{1-\frac9{10}}=\sqrt{\frac1{10}}=\dfrac1{\sqrt{10}}$, so $\tan\alpha=\dfrac{1/\sqrt{10}}{3/\sqrt{10}}=\dfrac13$.
   3. $\dfrac{\sqrt{1-\cos2\beta}}{\sqrt{1+\cos2\beta}} = \dfrac{\sqrt2\sin\beta}{\sqrt2\cos\beta}=\tan\beta = \dfrac17$.
   4. Compute $\tan2\alpha$ using the double-angle formula: $\tan2\alpha = \dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2/3}{1-1/9}=\dfrac{2/3}{8/9}=\dfrac34$. …

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3. 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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4. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If $7\cos\theta - \sin\theta = 5$ and $\tan\theta > 0$, then $\tan\theta =$ (A) $\dfrac{7}{12}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{4}{3}$ (D) $\dfrac{12}{7}$

   ›Reveal solutionSolution

   Solving $7\cos\theta-\sin\theta=5$ gives two possible $(\sin\theta,\cos\theta)$ pairs; the positivity condition on $\tan\theta$ picks out $\tan\theta=\dfrac34$.

   Concept and Intuition

   A linear equation in $\sin\theta$ and $\cos\theta$ combined with the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ generally has two solutions (geometrically, a line intersecting the unit circle in two points). Extra given conditions — here $\tan\theta>0$ — are exactly what's needed to select the physically/algebraically valid one.

   Step-by-Step Solution

   1. From $7\cos\theta-\sin\theta=5$, isolate $\sin\theta = 7\cos\theta - 5$.
   2. Substitute into $\sin^2\theta+\cos^2\theta=1$:

   $(7\cos\theta-5)^2+\cos^2\theta=1$

   3. Expand: $49\cos^2\theta-70\cos\theta+25+\cos^2\theta=1 \Rightarrow 50\cos^2\theta-70\cos\theta+24=0$.
   4. Divide by 2: $25\cos^2\theta-35\cos\theta+12=0$.
   5. Solve the quadratic: $\cos\theta=\dfrac{35\pm\sqrt{1225-1200}}{50}=\dfrac{35\pm5}{50}$, giving $\cos\theta=\dfrac45$ or $\cos\theta=\dfrac35$.
   6. Case $\cos\theta=\dfrac45$: $\sin\theta=7\left(\dfrac45\right)-5=\dfrac{28-25}{5}=\dfrac35$. Then $\tan\theta=\dfrac{3/5}{4/5}=\dfrac34>0$. ✓ …

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5. 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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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\sin 40^\circ \cos 80^\circ \left(\sec 260^\circ - \dfrac{1}{\sqrt{3}}\text{cosec } 280^\circ\right) =$ (A) $-\dfrac{1}{\sqrt{3}}$ (B) $\dfrac{1}{\sqrt{3}}$ (C) $\dfrac{4}{\sqrt{3}}$ (D) $4$

   ›Reveal solutionSolution

   Reducing $\sec260^\circ$ and $\csc280^\circ$ to reference angle $80^\circ$ (with correct signs) and simplifying the whole trigonometric expression gives $-\dfrac{1}{\sqrt3}$.

   Concept and Intuition

   The angles $260^\circ$ and $280^\circ$ aren't standard, but they're both close to $270^\circ$/$360^\circ$-type reference angles of $80^\circ$. Using $260^\circ=180^\circ+80^\circ$ (third quadrant, secant negative there) and $280^\circ=360^\circ-80^\circ$ (fourth quadrant, cosecant negative there) converts everything into functions of the single angle $80^\circ$, which then combines nicely with the given $\sin40^\circ\cos80^\circ$ factor.

   Step-by-Step Solution

   1. $\sec260^\circ = \sec(180^\circ+80^\circ) = -\sec80^\circ$ (secant is negative in the third quadrant).
   2. $\csc280^\circ = \csc(360^\circ-80^\circ) = -\csc80^\circ$ (cosecant is negative in the fourth quadrant).
   3. So $\sec260^\circ - \dfrac{1}{\sqrt3}\csc280^\circ = -\sec80^\circ + \dfrac{1}{\sqrt3}\csc80^\circ = \dfrac{-\sin80^\circ + \frac{1}{\sqrt3}\cos80^\circ}{\sin80^\circ\cos80^\circ}$.
   4. The whole expression becomes $\sin40^\circ\cos80^\circ\cdot\dfrac{-\sin80^\circ+\frac{1}{\sqrt3}\cos80^\circ}{\sin80^\circ\cos80^\circ} = \dfrac{\sin40^\circ}{\sin80^\circ}\left(-\sin80^\circ+\dfrac{1}{\sqrt3}\cos80^\circ\right)$. …

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7. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.If $0<x<\pi$ and $\cos x+\sin x=\dfrac{1}{2}$ then $\tan x =$ (A) $\dfrac{4-\sqrt7}{3}$ (B) $\dfrac{4+\sqrt7}{3}$ (C) $\dfrac{-(4+\sqrt7)}{3}$ (D) $\dfrac{-4+\sqrt7}{3}$

   ›Reveal solutionSolution

   Squaring the given sum turns it into a symmetric-function problem; picking the sign consistent with $0<x<\pi$ gives $\tan x = -\dfrac{4+\sqrt7}{3}$.

   Concept and Intuition

   Given $\sin x+\cos x$, squaring produces $\sin x\cos x$ via $(\sin x+\cos x)^2=1+2\sin x\cos x$. Once both the sum and product of $\sin x,\cos x$ are known, they are the two roots of a quadratic — solve it and use the domain restriction to pick which root is $\sin x$ and which is $\cos x$.

   Step-by-Step Solution

   1. $(\cos x+\sin x)^2 = \frac14 \Rightarrow 1+2\sin x\cos x=\frac14 \Rightarrow \sin x\cos x = -\frac38$.
   2. $\sin x,\cos x$ are roots of $t^2-\frac12 t-\frac38=0$, i.e. $8t^2-4t-3=0$.
   3. $t = \dfrac{4\pm\sqrt{16+96}}{16}=\dfrac{4\pm\sqrt{112}}{16}=\dfrac{4\pm4\sqrt7}{16}=\dfrac{1\pm\sqrt7}{4}$.
   4. Since $0<x<\pi$, $\sin x>0$. Since $\sin x\cos x=-\frac38<0$, $\cos x$ must be negative. So $\sin x=\dfrac{1+\sqrt7}{4}$ (positive) and $\cos x=\dfrac{1-\sqrt7}{4}$ (negative, as $\sqrt7>1$). …

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8. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
   Q.The value of $\cos\left(\dfrac{7\pi}{12}\right)$ is (A) $\dfrac{\sqrt2+\sqrt3}{4}$ (B) $\dfrac{\sqrt2-\sqrt3}{4}$ (C) $\dfrac{\sqrt2-\sqrt6}{4}$ (D) $\dfrac{\sqrt2+\sqrt6}{4}$

   ›Reveal solutionSolution

   This tests the compound-angle formula for cosine applied to a non-standard angle written as a sum of two standard angles.

   Concept and Intuition

   $7\pi/12$ radians $=105°$ isn't a standard angle by itself, but it splits nicely as $60°+45°$, both standard angles whose sine/cosine values are known exactly.

   Step-by-Step Solution

   1. $\cos(105°) = \cos(60°+45°) = \cos60°\cos45°-\sin60°\sin45°$.
   2. Substitute values: $\cos60°=\tfrac12,\ \cos45°=\tfrac{\sqrt2}{2},\ \sin60°=\tfrac{\sqrt3}{2},\ \sin45°=\tfrac{\sqrt2}{2}$. …

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9. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.$\cos^2 10° + \cos^2 50° - \sin 40° \sin 80° =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{4}{3}$ (D) $\dfrac{3}{4}$

   ›Reveal solutionSolution

   Using the power-reduction and product-to-sum identities, all the $\cos40°$ terms cancel between the two parts of the expression, leaving a clean numeric value of $3/4$.

   Concept and Intuition

   Expressions combining squared cosines and a product of sines can be systematically simplified using two standard identities: the power-reduction formula $\cos^2\theta=\frac{1+\cos2\theta}{2}$ (to remove the squares) and the product-to-sum formula $\sin A\sin B = \frac12[\cos(A-B)-\cos(A+B)]$ (to remove the product). Doing this carefully often causes matching cosine terms to cancel, leaving a pure number.

   Step-by-Step Solution

   1. Apply the power-reduction formula to each squared cosine: $\cos^210° = \dfrac{1+\cos20°}{2}$ and $\cos^250° = \dfrac{1+\cos100°}{2}$.
   2. Sum them: $\cos^210°+\cos^250° = 1 + \dfrac{\cos20°+\cos100°}{2}$.
   3. Simplify $\cos20°+\cos100°$ using sum-to-product: $\cos A+\cos B = 2\cos\left(\dfrac{A+B}{2}\right)\cos\left(\dfrac{A-B}{2}\right)$, with $A=20°,B=100°$: $2\cos(60°)\cos(-40°) = 2\times\tfrac12\times\cos40° = \cos40°$ (using $\cos(-40°)=\cos40°$).
   4. So $\cos^210°+\cos^250° = 1+\dfrac{\cos40°}{2}$.
   5. Now simplify $\sin40°\sin80°$ using product-to-sum: $\sin A\sin B = \dfrac12[\cos(A-B)-\cos(A+B)]$, with $A=40°,B=80°$: $\dfrac12[\cos(-40°)-\cos(120°)] = \dfrac12[\cos40°-(-\tfrac12)] = \dfrac{\cos40°}{2}+\dfrac14$. …

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10. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.$\tan9^{\circ}-\tan27^{\circ}-\tan63^{\circ}+\tan81^{\circ}=$ (A) 4 (B) 3 (C) 2 (D) 1

    ›Reveal solutionSolution

    This uses the co-function identity $\tan(90°-\theta)=\cot\theta$ to pair up terms, then the exact values of $\sin18°$ and $\sin54°$; the result is exactly $4$.

    Concept and Intuition

    $81°=90°-9°$ and $63°=90°-27°$, so $\tan81°=\cot9°$ and $\tan63°=\cot27°$. Regrouping the four terms into two "$\tan+\cot$" pairs turns the expression into a difference of two $\dfrac{2}{\sin(2\theta)}$-type quantities, which can be evaluated exactly using the known closed forms of $\sin18°$ and $\sin54°$ (related to the golden ratio).

    Step-by-Step Solution

    1. Rewrite: $\tan9°-\tan27°-\tan63°+\tan81° = (\tan9°+\tan81°)-(\tan27°+\tan63°) = (\tan9°+\cot9°)-(\tan27°+\cot27°)$.
    2. Use $\tan\theta+\cot\theta=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}=\dfrac{2}{\sin2\theta}$: so the expression becomes $\dfrac{2}{\sin18°}-\dfrac{2}{\sin54°}$.
    3. Exact values: $\sin18°=\dfrac{\sqrt5-1}{4}$, $\sin54°=\cos36°=\dfrac{\sqrt5+1}{4}$. …

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11. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.If $\sin\left(\dfrac{\pi}{4}\cot\theta\right) = \cos\left(\dfrac{\pi}{4}\tan\theta\right)$, then $\theta =$ (A) $2n\pi + \dfrac{\pi}{4}$ (B) $2n\pi \pm \dfrac{\pi}{4}$ (C) $2n\pi - \dfrac{\pi}{4}$ (D) $n\pi + \dfrac{\pi}{4}$

    ›Reveal solutionSolution

    Converting cosine to sine of the complementary angle and testing candidate solutions shows the general solution repeats every $\pi$, giving $\theta=n\pi+\dfrac{\pi}{4}$.

    Concept and Intuition

    Both $\tan\theta$ and $\cot\theta$ have period $\pi$, so any solution set for an equation built purely from them should also have period $\pi$ (not $2\pi$). This immediately makes options with a $2n\pi$ structure suspicious, and testing values confirms which option is right.

    Step-by-Step Solution

    1. Rewrite the RHS using $\cos x=\sin\left(\dfrac{\pi}{2}-x\right)$: the equation becomes $\sin\left(\dfrac{\pi}{4}\cot\theta\right)=\sin\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\tan\theta\right)$.
    2. A natural guess is $\cot\theta=\tan\theta$, i.e. $\tan^2\theta=1 \Rightarrow \tan\theta=\pm1$, giving $\theta=\dfrac{\pi}{4}$ as a base solution (then both sides read $\sin(\pi/4)=\sqrt2/2$).
    3. Since $\tan\theta$ and $\cot\theta$ each repeat every $\pi$, check $\theta=\dfrac{\pi}{4}+\pi=\dfrac{5\pi}{4}$: $\tan\theta=1,\cot\theta=1$ again, so the equation again holds. …

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12. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.$\cos^4\dfrac{\pi}{12} + \cos^4\dfrac{5\pi}{12} + \cos^4\dfrac{7\pi}{12} + \cos^4\dfrac{11\pi}{12} =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{5}{3}$ (D) $\dfrac{7}{4}$

    ›Reveal solutionSolution

    Pairing the four angles by supplementary-angle and complementary-angle symmetry collapses the sum to $2(\sin^4\theta+\cos^4\theta)$ at $\theta=\pi/12$, giving $\dfrac74$.

    Concept and Intuition

    The four angles $\pi/12, 5\pi/12, 7\pi/12, 11\pi/12$ aren't independent — they come in supplementary pairs ($\pi/12$ & $11\pi/12$; $5\pi/12$ & $7\pi/12$), and $\cos(\pi-\theta)=-\cos\theta$, so raising to an even power ($4$th) erases the sign difference entirely. That collapses four terms into two duplicated pairs. Then noticing $5\pi/12=\pi/2-\pi/12$ turns one of those cosines into a sine of the same angle as the other, so the whole sum reduces to the well-known identity for $\sin^4\theta+\cos^4\theta$.

    Step-by-Step Solution

    1. $\dfrac{7\pi}{12}=\pi-\dfrac{5\pi}{12}$, so $\cos\dfrac{7\pi}{12}=-\cos\dfrac{5\pi}{12}$, and since the power is even, $\cos^4\dfrac{7\pi}{12}=\cos^4\dfrac{5\pi}{12}$.
    2. $\dfrac{11\pi}{12}=\pi-\dfrac{\pi}{12}$, so similarly $\cos^4\dfrac{11\pi}{12}=\cos^4\dfrac{\pi}{12}$.
    3. Sum $= 2\cos^4\dfrac{\pi}{12}+2\cos^4\dfrac{5\pi}{12}$.
    4. $\dfrac{5\pi}{12}=\dfrac{\pi}{2}-\dfrac{\pi}{12}$, so $\cos\dfrac{5\pi}{12}=\sin\dfrac{\pi}{12}$; let $\theta=\pi/12$. Sum $=2\cos^4\theta+2\sin^4\theta = 2(\sin^4\theta+\cos^4\theta)$.
    5. Use $\sin^4\theta+\cos^4\theta = 1-2\sin^2\theta\cos^2\theta = 1-\dfrac12\sin^2(2\theta)$. …

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