Q.Show that $\tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right)=\frac{4-\sqrt7}{3}$ and justify why the other value $\frac{4+\sqrt7}{3}$ is ignored.

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1. 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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2. 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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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 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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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 2024Set eng-2024-05-20-AN1 markMCQ
   Q.If $(\alpha + \beta)$ is not a multiple of $\dfrac{\pi}{2}$ and $3\sin(\alpha - \beta) = 5\cos(\alpha + \beta)$, then $\tan\left(\dfrac{\pi}{4} + \alpha\right) + 4\tan\left(\dfrac{\pi}{4} + \beta\right) =$ (A) $0$ (B) $1$ (C) $4$ (D) $2$

   ›Reveal solutionSolution

   This tests converting a trig equation into a tan-relation and recognising that the target expression is forced to a constant value by the given constraint. The answer is $0$.

   Concept and Intuition

   Whenever a trig condition mixes $\sin(\alpha-\beta)$ and $\cos(\alpha+\beta)$, dividing through by $\cos\alpha\cos\beta$ converts everything into $\tan\alpha,\tan\beta$ — this is the standard bridge to compound-angle tangent identities like $\tan(\pi/4+\theta)=\dfrac{1+\tan\theta}{1-\tan\theta}$.

   Step-by-Step Solution

   1. Expand: $3\sin(\alpha-\beta) = 3(\sin\alpha\cos\beta-\cos\alpha\sin\beta)$ and $5\cos(\alpha+\beta)=5(\cos\alpha\cos\beta-\sin\alpha\sin\beta)$.
   2. So $3\sin\alpha\cos\beta-3\cos\alpha\sin\beta = 5\cos\alpha\cos\beta-5\sin\alpha\sin\beta$.
   3. Divide throughout by $\cos\alpha\cos\beta$ (valid since $\alpha+\beta$ isn't an odd multiple of $\pi/2$, so neither cosine vanishes generically here): letting $x=\tan\alpha,\ y=\tan\beta$, we get $3x - 3y = 5 - 5xy$, i.e. $3x-3y+5xy-5=0$.
   4. We want $S=\dfrac{1+x}{1-x}+4\cdot\dfrac{1+y}{1-y}$. Take a convenient solution of the constraint: set $y=0$. Then $3x-5=0\Rightarrow x=5/3$.
   5. $S = \dfrac{1+5/3}{1-5/3} + 4(1) = \dfrac{8/3}{-2/3}+4 = -4+4=0$. …

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7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.Determine the value of 'a' in $\tan 70^\circ - \tan 20^\circ = a\cdot \tan 50^\circ$? (A) -4 (B) 4 (C) -2 (D) 2

   ›Reveal solutionSolution

   Rewriting $\tan70^\circ$ as $\cot20^\circ$ turns the difference into the double-angle cotangent identity, giving $a=2$.

   Concept and Intuition

   Complementary-angle identities ($\tan(90^\circ-\theta)=\cot\theta$) combined with the double-angle formula for $\cot2\theta$ let a seemingly awkward difference of tangents collapse to a single tangent term.

   Step-by-Step Solution

   1. $\tan70^\circ=\tan(90^\circ-20^\circ)=\cot20^\circ=\dfrac{1}{\tan20^\circ}$.
   2. So $\tan70^\circ-\tan20^\circ=\dfrac{1}{\tan20^\circ}-\tan20^\circ=\dfrac{1-\tan^220^\circ}{\tan20^\circ}$.
   3. Recall $\cot2\theta=\dfrac{1-\tan^2\theta}{2\tan\theta}$, so $\dfrac{1-\tan^2\theta}{\tan\theta}=2\cot2\theta$. With $\theta=20^\circ$: expression $=2\cot40^\circ$.
   4. $\cot40^\circ=\tan(90^\circ-40^\circ)=\tan50^\circ$.
   5. So $\tan70^\circ-\tan20^\circ=2\tan50^\circ$, giving $a=2$. …

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8. 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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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 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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11. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If $1+\sqrt{1+a}=(1+\sqrt{1-a})\cot\alpha$ and $0<a<1$, then $\sin 4\alpha=$ (A) $a$ (B) $2a$ (C) $3a$ (D) $4a$

    ›Reveal solutionSolution

    A clever substitution $a=\sin2\theta$ converts the given surd equation into a clean tangent-addition identity, showing directly that $\sin4\alpha=a$.

    Concept and Intuition

    Expressions like $\sqrt{1\pm\sin2\theta}$ simplify beautifully to $|\cos\theta\pm\sin\theta|$ because $1\pm\sin2\theta=(\sin\theta\pm\cos\theta)^2$. This is the standard trick for equations dressed up with nested square roots of $1\pm a$.

    Step-by-Step Solution

    1. Since $0<a<1$, write $a=\sin2\theta$ with $\theta\in(0,\pi/4)$.
    2. $1+a=1+\sin2\theta=(\sin\theta+\cos\theta)^2\Rightarrow\sqrt{1+a}=\sin\theta+\cos\theta$.
    3. $1-a=1-\sin2\theta=(\cos\theta-\sin\theta)^2\Rightarrow\sqrt{1-a}=\cos\theta-\sin\theta$ (positive since $\theta<\pi/4$).
    4. Given equation: $1+\sin\theta+\cos\theta=(1+\cos\theta-\sin\theta)\cot\alpha$.
    5. Using half-angle forms $1+\cos\theta=2\cos^2(\theta/2)$ and $\sin\theta=2\sin(\theta/2)\cos(\theta/2)$: $\cot\alpha=\dfrac{2\cos(\theta/2)[\cos(\theta/2)+\sin(\theta/2)]}{2\cos(\theta/2)[\cos(\theta/2)-\sin(\theta/2)]}=\dfrac{1+\tan(\theta/2)}{1-\tan(\theta/2)}=\tan\!\left(\dfrac\pi4+\dfrac\theta2\right).$ …

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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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