Q.Prove that $\cot\left(\frac{\pi}{4}-2\cot^{-1}3\right)=7$.

Showing the 12 most recent of 146 on this concept.

1. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.$\cot16^{\circ}\cot44^{\circ}+\cot44^{\circ}\cot76^{\circ}-\cot76^{\circ}\cot16^{\circ}=$ (A) 1 (B) $-1$ (C) $-3$ (D) 3

   ›Reveal solutionSolution

   A three-cotangent-product combination evaluates to a clean constant; direct numerical evaluation confirms it. Answer: 3.

   Concept and Intuition

   Expressions of the form $\sum\cot A\cot B$ over three angles often simplify to a constant when the angles carry a hidden special relationship (as with the classic $\tan A\tan B+\tan B\tan C+\tan C\tan A=1$ identity for $A+B+C=90^\circ$). Here the safest and fastest route for a numeric MCQ is to evaluate directly and match to the given options, since all four options are simple integers.

   Step-by-Step Solution

   1. Compute the three cotangents: $\cot16^\circ\approx3.48741$, $\cot44^\circ\approx1.03553$, $\cot76^\circ\approx0.24933$.
   2. $\cot16^\circ\cot44^\circ\approx3.48741\times1.03553\approx3.61139$.
   3. $\cot44^\circ\cot76^\circ\approx1.03553\times0.24933\approx0.25822$.
   4. $\cot76^\circ\cot16^\circ\approx0.24933\times3.48741\approx0.86954$.
   5. Sum with the given signs: $3.61139+0.25822-0.86954\approx3.00007\approx3$.

   Common Mistakes

   • Sign error on the last term (it is subtracted, not added).
   • Assuming a simpler-looking identity applies without checking that the angles actually satisfy its precondition.

   ✓Final answer

   The correct option is (D) — 3.

   ANSWER: D

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: $\tan2\alpha = \dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2/3}{1-1/9}=\dfrac{2/3}{8/9}=\dfrac34$.
   5. Compute $\tan(2\alpha+\beta)$: $\dfrac{\tan2\alpha+\tan\beta}{1-\tan2\alpha\tan\beta} = \dfrac{3/4+1/7}{1-(3/4)(1/7)} = \dfrac{21/28+4/28}{1-3/28} = \dfrac{25/28}{25/28}=1$.
   6. Since $\tan(2\alpha+\beta)=1$, and $2\alpha+\beta$ lies strictly between $0$ and $2\cdot\frac\pi4+\frac\pi4=\frac{3\pi}4$ (because $0<\alpha<\pi/4\Rightarrow0<2\alpha<\pi/2$, and $0<\beta<\pi/4$), the only angle in $(0,3\pi/4)$ with tangent $1$ is $\dfrac\pi4$.

   Common Mistakes

   • Forgetting to restrict the general solution $2\alpha+\beta=\pi/4+k\pi$ to the actual valid range implied by the given bounds on $\alpha,\beta$ — this is what rules out $5\pi/4$ etc.

   ✓Final answer

   The correct option is (D) — $\dfrac\pi4$.

   ANSWER: D

3. 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 $\tan\theta=\dfrac{3/5}{4/5}=\dfrac34>0$. ✓
   7. Case $\cos\theta=\dfrac35$: $\sin\theta=7\left(\dfrac35\right)-5=\dfrac{21-25}{5}=-\dfrac45$. Then $\tan\theta=\dfrac{-4/5}{3/5}=-\dfrac43<0$. Rejected.
   8. So $\tan\theta=\dfrac34$.

   Common Mistakes

   • Forgetting to check the sign condition $\tan\theta>0$ and reporting both roots or the wrong one.
   • Squaring introduces an extraneous root — always verify each root against the original linear equation and the given condition.

   ✓Final answer

   The correct option is (B) — $\dfrac{3}{4}$.

   ANSWER: B

4. 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: $\dfrac{4\cos4A\cos2A\cos A}{4\sin4A\cos2A\cos A} = \dfrac{\cos4A}{\sin4A} = \cot4A$.
   4. With $A=\dfrac{\pi}{24}$: $4A = \dfrac{4\pi}{24}=\dfrac{\pi}{6}=30^\circ$.
   5. $\cot30^\circ = \sqrt3$.

   Common Mistakes

   • Trying to expand each term individually with multiple-angle formulas rather than pairing symmetric terms — far more error-prone.
   • Forgetting that $\cos2A\cos A$ cancels exactly between numerator and denominator only when both are grouped identically.

   ✓Final answer

   The correct option is (A) — $\sqrt{3}$.

   ANSWER: A

5. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.$1 + \cot^2 30^0 - \sec^2 45^0 =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1-\sqrt{3}}{2}$ (C) $2$ (D) $0$

   ›Reveal solutionSolution

   Direct substitution of standard angle values gives $1+3-2=2$.

   Concept and Intuition

   This is a pure standard-angle evaluation: $\cot30°=\sqrt3$ and $\sec45°=\sqrt2$ are memorized exact values.

   Step-by-Step Solution

   1. $\cot30° = \sqrt3 \Rightarrow \cot^230°=3$.
   2. $\sec45° = \sqrt2 \Rightarrow \sec^245°=2$.
   3. Substitute: $1+3-2=2$.

   Common Mistakes

   • Confusing $\cot30°$ with $\tan30°=1/\sqrt3$.
   • Sign error combining the three terms.

   ✓Final answer

   The correct option is (C) — $2$.

   ANSWER: C

6. 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 $\theta=\pi/4$): $\tan\frac{\pi}{8} = \tan22.5^\circ = \sqrt2-1$.
   4. So $\tan\frac{7\pi}{8} = -(\sqrt2-1) = 1-\sqrt2$.

   Common Mistakes

   • Forgetting the negative sign from the second-quadrant reduction.
   • Misremembering $\tan\frac{\pi}{8}$ as $\sqrt2+1$ (that is $\cot\frac{\pi}{8}$ or $\tan\frac{3\pi}{8}$).

   ✓Final answer

   The correct option is (B) — $1-\sqrt2$.

   ANSWER: B

7. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $1-\cot 23^0 = \dfrac{x}{1-\cot 22^0}$, then x = (A) 1 (B) 2 (C) $\dfrac{1}{2}$ (D) 3

   ›Reveal solutionSolution

   Recognizing $23^\circ+22^\circ=45^\circ$ triggers a standard cotangent identity that instantly evaluates the product without needing actual trig values.

   Concept and Intuition

   Whenever $A+B=45^\circ$, the identity $(\cot A-1)(\cot B-1)=2$ holds — this follows directly from the cotangent addition formula $\cot(A+B)=\dfrac{\cot A\cot B-1}{\cot A+\cot B}=1$.

   Step-by-Step Solution

   1. Rearrange the given equation: $x=(1-\cot23^\circ)(1-\cot22^\circ)$.
   2. Note $23^\circ+22^\circ=45^\circ$.
   3. From $\cot(A+B)=1$ when $A+B=45^\circ$: $\dfrac{\cot A\cot B-1}{\cot A+\cot B}=1 \Rightarrow \cot A\cot B-1=\cot A+\cot B$.
   4. So $\cot A\cot B-\cot A-\cot B+1 = (\cot A+\cot B+1)-\cot A-\cot B+1=2$, i.e. $(\cot A-1)(\cot B-1)=2$.
   5. Since $(1-\cot23^\circ)(1-\cot22^\circ)=(-(\cot23^\circ-1))(-(\cot22^\circ-1))=(\cot23^\circ-1)(\cot22^\circ-1)=2$.
   6. Hence $x=2$.

   Common Mistakes

   • Trying to look up or approximate $\cot23^\circ$ and $\cot22^\circ$ numerically instead of using the identity.
   • Sign error not noticing that $(1-\cot A)(1-\cot B)=(\cot A-1)(\cot B-1)$ (double negative cancels).

   ✓Final answer

   The correct option is (B) — 2.

   ANSWER: B

8. 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 $\sqrt7>1$).
   5. $\tan x = \dfrac{1+\sqrt7}{1-\sqrt7}$. Rationalize: multiply by $\dfrac{1+\sqrt7}{1+\sqrt7}$... instead multiply num & denom by $(1+\sqrt7)$: numerator $(1+\sqrt7)^2=8+2\sqrt7$, denominator $(1-\sqrt7)(1+\sqrt7)=1-7=-6$.
   6. $\tan x = \dfrac{8+2\sqrt7}{-6} = -\dfrac{4+\sqrt7}{3}$.

   Common Mistakes

   • Picking the wrong root for $\sin x$ vs $\cos x$ (must use the sign of the product plus the domain restriction).
   • Sign errors while rationalizing the denominator.

   ✓Final answer

   The correct option is (C) — $\dfrac{-(4+\sqrt7)}{3}$.

   ANSWER: C

9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$2\cot^2\theta - \cot\theta - 3 =$ (A) $(2\cot\theta - 3)(\cot\theta + 1)$ (B) $(2\cot\theta - 1)(\cot\theta + 3)$ (C) $(2\cot\theta + 3)(\cot\theta - 1)$ (D) $(2\cot\theta + 1)(\cot\theta - 3)$

   ›Reveal solutionSolution

   A plain quadratic factorisation in $\cot\theta$: $2\cot^2\theta-\cot\theta-3=(2\cot\theta-3)(\cot\theta+1)$.

   Concept and Intuition

   Trig "identities" that are really just quadratics in disguise are best handled by substituting a

   single variable for the trig function, factoring as ordinary algebra, then substituting back.

   Step-by-Step Solution

   1. Let $c=\cot\theta$. The expression is $2c^2-c-3$.
   2. Find factors of $2\times(-3)=-6$ that sum to the middle coefficient $-1$: these are $2$ and $-3$.
   3. Split the middle term: $2c^2+2c-3c-3 = 2c(c+1)-3(c+1) = (2c-3)(c+1)$.
   4. Substitute back: $(2\cot\theta-3)(\cot\theta+1)$.
   5. Verify by expanding: $2\cot^2\theta+2\cot\theta-3\cot\theta-3 = 2\cot^2\theta-\cot\theta-3$. Matches.

   Common Mistakes

   • Sign slip giving $(2\cot\theta+3)(\cot\theta-1)$, which expands to $2\cot^2\theta+\cot\theta-3$ (wrong middle sign).
   • Forgetting to verify the factorisation by re-expanding.

   ✓Final answer

   The correct option is (A) — $(2\cot\theta-3)(\cot\theta+1)$.

   ANSWER: A

10. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.If $f(x) = \dfrac{\cot x}{1+\cot x}$ and $\alpha + \beta = \dfrac{5\pi}{4}$ then the value of $f(\alpha)f(\beta) =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{-3}{2}$ (C) $\dfrac{-1}{2}$ (D) $\dfrac{1}{2}$

    ›Reveal solutionSolution

    Rewriting $f(x)$ as $\dfrac{\cos x}{\sin x+\cos x}$ and using $\alpha+\beta=5\pi/4$, the product $f(\alpha)f(\beta)$ evaluates to the constant $\dfrac12$.

    Concept and Intuition

    $f(x)=\dfrac{\cot x}{1+\cot x}$ simplifies by multiplying through by $\sin x$ to $\dfrac{\cos x}{\sin x+\cos x}$, which is a much friendlier form for evaluating at complementary-type angle pairs.

    Step-by-Step Solution

    1. $f(x)=\dfrac{\cos x/\sin x}{1+\cos x/\sin x}=\dfrac{\cos x}{\sin x+\cos x}$.
    2. Pick a convenient pair with $\alpha+\beta=5\pi/4$: let $\alpha=\pi/3$, so $\beta=5\pi/4-\pi/3=11\pi/12$.
    3. $f(\pi/3)=\dfrac{\cos60^\circ... }{}$; concretely $\cos(\pi/3)=1/2,\sin(\pi/3)=\sqrt3/2$, so $f(\pi/3)=\dfrac{1/2}{(\sqrt3+1)/2}=\dfrac1{\sqrt3+1}=\dfrac{\sqrt3-1}{2}$.
    4. $\cos(11\pi/12)=-\cos15^\circ=-\dfrac{\sqrt6+\sqrt2}{4}$, $\sin(11\pi/12)=\sin15^\circ=\dfrac{\sqrt6-\sqrt2}{4}$; their sum is $-\dfrac{\sqrt2}{2}$, giving $f(11\pi/12)=\dfrac{-(\sqrt6+\sqrt2)/4}{-\sqrt2/2}=\dfrac{\sqrt3+1}{2}$.
    5. Product: $\dfrac{\sqrt3-1}{2}\cdot\dfrac{\sqrt3+1}{2}=\dfrac{3-1}{4}=\dfrac12$. This value is the same for any valid $\alpha$ satisfying the sum constraint (a general algebraic proof confirms it's angle-independent).

    Common Mistakes

    • Assuming the answer depends on which specific $\alpha,\beta$ pair is picked — the constraint $\alpha+\beta=5\pi/4$ makes the product a genuine constant.

    ✓Final answer

    The correct option is (D) — $\dfrac{1}{2}$.

    ANSWER: D

11. 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: $\cos60°=\tfrac12,\ \cos45°=\tfrac{\sqrt2}{2},\ \sin60°=\tfrac{\sqrt3}{2},\ \sin45°=\tfrac{\sqrt2}{2}$.
    3. $=\tfrac12\cdot\tfrac{\sqrt2}{2} - \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt2}{2} = \tfrac{\sqrt2}{4}-\tfrac{\sqrt6}{4} = \dfrac{\sqrt2-\sqrt6}{4}$.

    Common Mistakes

    • Splitting as $45°+60°$ but mixing up a sign (using $+$ instead of $-$) — remember $\cos(A+B)$ has a minus sign in its expansion.

    ✓Final answer

    The correct option is (C) — $\dfrac{\sqrt2-\sqrt6}{4}$.

    ANSWER: C

12. 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. $\cos\theta + \cos3\theta = 2\cos\left(\dfrac{\theta+3\theta}{2}\right)\cos\left(\dfrac{3\theta-\theta}{2}\right) = 2\cos2\theta\cos\theta$.
    3. Ratio: $\dfrac{2\sin2\theta\cos\theta}{2\cos2\theta\cos\theta} = \dfrac{\sin2\theta}{\cos2\theta} = \tan2\theta$ (the $\cos\theta$ factors cancel, assuming $\cos\theta\ne0$).

    Common Mistakes

    • Mixing up the sum-to-product formulas for sine and cosine (their "outer" trig function differs: sin+sin gives sin·cos, cos+cos gives cos·cos).
    • Forgetting the common factor cancels cleanly, leading to an unnecessarily complicated final expression.

    ✓Final answer

    The correct option is (C) — $\tan2\theta$.

    ANSWER: C