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.

Showing the 12 most recent of 34 on this concept.

1. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.$2 \tan^{-1} \left( \dfrac{1}{3} \right) + \tan^{-1} \left( \dfrac{1}{7} \right) =$ (A) $\tan^{-1} \left( \dfrac{49}{29} \right)$ (B) $\dfrac{\pi}{2}$ (C) $0$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   We simplify the expression by first converting $2 \tan^{-1} \left( \frac{1}{3} \right)$ into a single $\tan^{-1}$ term, then combining it with $\tan^{-1} \left( \frac{1}{7} \right)$ using the sum formula for inverse tangents. The final result is $\boxed{\frac{\pi}{4}}$.

   The problem asks us to evaluate an expression involving inverse tangent functions. The key to solving this is to use the standard addition formulas for inverse tangents to simplify the expression step-by-step. We have a term of the form $2 \tan^{-1} x$ and then a sum of two $\tan^{-1}$ terms.

   Here are the relevant formulas we will use:

   $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$, for $-1 < x < 1$.

   [!FORMULA]

   $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$, for $xy < 1$.

   Let's break down the calculation.

   1. Simplify the $2 \tan^{-1} \left( \frac{1}{3} \right)$ term: We start by simplifying the first part of the expression, $2 \tan^{-1} \left( \frac{1}{3} \right)$. We use the formula $2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1-x^2} \right)$. Here, $x = \frac{1}{3}$. Since $-1 < \frac{1}{3} < 1$, the formula is applicable.

   $2 \tan^{-1} \left( \frac{1}{3} \right) = \tan^{-1} \left( \frac{2 \left( \frac{1}{3} \right)}{1 - \left( \frac{1}{3} \right)^2} \right)$

   $= \tan^{-1} \left( \frac{\frac{2}{3}}{1 - \frac{1}{9}} \right)$

   $= \tan^{-1} \left( \frac{\frac{2}{3}}{\frac{9-1}{9}} \right)$

   $= \tan^{-1} \left( \frac{\frac{2}{3}}{\frac{8}{9}} \right)$

   To simplify the fraction, we multiply the numerator by the reciprocal of the denominator:
   

   $= \tan^{-1} \left( \frac{2}{3} \times \frac{9}{8} \right)$

   $= \tan^{-1} \left( \frac{18}{24} \right)$

   $= \tan^{-1} \left( \frac{3}{4} \right)$

   So, the original expression becomes $\tan^{-1} \left( \frac{3}{4} \right) + \tan^{-1} \left( \frac{1}{7} \right)$.
   

   2. Combine the two $\tan^{-1}$ terms:

   Now we have an expression of the form $\tan^{-1} x + \tan^{-1} y$, where $x = \frac{3}{4}$ and $y = \frac{1}{7}$. We use the formula $\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$.

   First, we check the condition $xy < 1$:

   $xy = \left( \frac{3}{4} \right) \left( \frac{1}{7} \right) = \frac{3}{28}$. Since $\frac{3}{28} < 1$, the formula is applicable. …

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2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\tan^{-1} \dfrac{1}{2} + \tan^{-1} \dfrac{1}{3} + \tan^{-1} \dfrac{2}{3} + \tan^{-1} \dfrac{1}{5} =$ (A) $\dfrac{\pi}{4}$ (B) $\tan^{-1} \left( \dfrac{7}{11} \right)$ (C) $\dfrac{\pi}{2}$ (D) $\tan^{-1} \left( \dfrac{23}{24} \right)$

   ›Reveal solutionSolution

   The sum of the four arctangents simplifies to $\pi/4$ by repeatedly applying the tangent addition formula and noticing that the combined angle lies in the first quadrant. The correct option is (A).

   We are asked to evaluate

   $\tan^{-1}\frac12 + \tan^{-1}\frac13 + \tan^{-1}\frac23 + \tan^{-1}\frac15.$

   The key idea is to combine arctangents two at a time using the formula

   $\tan^{-1}a + \tan^{-1}b = \tan^{-1}\frac{a+b}{1-ab},$

   but we must always check the quadrant of the resulting angle. Since all given fractions are positive and less than 1, each arctangent lies in $(0,\pi/4)$. Their sum will be less than $\pi$, so we can safely use the formula without worrying about adding $\pi$ corrections.

   1. Combine the first two terms Let $\alpha = \tan^{-1}\frac12$ and $\beta = \tan^{-1}\frac13$. Then

   $\tan(\alpha+\beta) = \frac{\frac12+\frac13}{1-\frac12\cdot\frac13} = \frac{\frac{5}{6}}{1-\frac16} = \frac{5/6}{5/6} = 1.$

   Since $\alpha,\beta < \pi/4$, their sum is less than $\pi/2$ and positive, so

   $\alpha+\beta = \tan^{-1}1 = \frac{\pi}{4}.$

   2. Combine the next two terms Let $\gamma = \tan^{-1}\frac23$ and $\delta = \tan^{-1}\frac15$. Then

   $\tan(\gamma+\delta) = \frac{\frac23+\frac15}{1-\frac23\cdot\frac15} = \frac{\frac{10}{15}+\frac{3}{15}}{1-\frac{2}{15}} = \frac{13/15}{13/15} = 1.$

   Again, both angles are less than $\pi/4$, so their sum is also $\pi/4$.

   3. Add the two results Now we have

   $\left(\tan^{-1}\frac12+\tan^{-1}\frac13\right) + \left(\tan^{-1}\frac23+\tan^{-1}\frac15\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}.$

   That would suggest the answer is $\pi/2$, but wait — we must check if the sum of all four original angles really equals $\pi/2$ or if there is a subtlety.

   Watch out

   The sum of two arctangents each equal to $\pi/4$ is $\pi/2$, but the original four angles are all less than $\pi/4$, so their total is less than $\pi$. However, $\pi/2$ is a valid possibility. But let’s verify by combining all four at once to be safe.

   4. Combine all four directly Let $S = \tan^{-1}\frac12 + \tan^{-1}\frac13 + \tan^{-1}\frac23 + \tan^{-1}\frac15$. …

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3. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If $\cos x + \cos y = \dfrac{2}{3}$ and $\sin x - \sin y = \dfrac{3}{4}$, then $\sin(x-y) + \cos(x-y) =$ (A) $\dfrac{161}{145}$ (B) $\dfrac{127}{145}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{8}{9}$

   ›Reveal solutionSolution

   We use sum‑to‑product identities to find $\cos\frac{x+y}{2}$ and $\sin\frac{x+y}{2}$, then compute $\sin(x-y)$ and $\cos(x-y)$ via double‑angle formulas, obtaining $\frac{161}{145}$.

   Concept & Intuition

   We are given two equations mixing sums and differences of sines and cosines. The classic trick is to rewrite each as a product using sum‑to‑product identities. That isolates the half‑sum and half‑difference angles. Then we can find $\sin\frac{x-y}{2}$ and $\cos\frac{x-y}{2}$ from the given numbers, and finally use double‑angle formulas to get $\sin(x-y)$ and $\cos(x-y)$.

   ---

   1. Apply sum‑to‑product identities

   $\cos x + \cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2} = \frac{2}{3}$

   $\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2} = \frac{3}{4}$

   2. Divide the two equations to eliminate $\cos\frac{x+y}{2}$ (provided it is nonzero):

   $\frac{2\cos\frac{x+y}{2}\sin\frac{x-y}{2}}{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}} = \frac{3/4}{2/3}$

   $\tan\frac{x-y}{2} = \frac{3}{4} \cdot \frac{3}{2} = \frac{9}{8}$

   3. Find $\sin\frac{x-y}{2}$ and $\cos\frac{x-y}{2}$ from the tangent. Let $t = \frac{x-y}{2}$. Then $\tan t = \frac{9}{8}$. Construct a right triangle: opposite = 9, adjacent = 8, hypotenuse = $\sqrt{9^2+8^2} = \sqrt{145}$. Hence

   $\sin t = \frac{9}{\sqrt{145}}, \qquad \cos t = \frac{8}{\sqrt{145}}.$

   4. Use double‑angle formulas to get $\sin(x-y)$ and $\cos(x-y)$: $$$$

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4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\tan^{-1} \dfrac{3}{5} + \tan^{-1} \dfrac{6}{41} + \tan^{-1} \dfrac{9}{191} =$ (A) $\tan^{-1} \dfrac{9}{10}$ (B) $\tan^{-1} \dfrac{18}{19}$ (C) $\tan^{-1} \dfrac{3}{191}$ (D) $\tan^{-1} \dfrac{6}{205}$

   ›Reveal solutionSolution

   Adding the arctangents two at a time gives $\tan^{-1}\dfrac{9}{10}$.

   Solution

   Use $\tan^{-1}p+\tan^{-1}q=\tan^{-1}\dfrac{p+q}{1-pq}$.

   First two terms:

   $\tan^{-1}\frac{3}{5}+\tan^{-1}\frac{6}{41}=\tan^{-1}\frac{\frac{3}{5}+\frac{6}{41}}{1-\frac{3}{5}\cdot\frac{6}{41}}=\tan^{-1}\frac{123+30}{205-18}=\tan^{-1}\frac{153}{187}=\tan^{-1}\frac{9}{11}.$

   Add the third term: …

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5. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.$\cos^{-1} \dfrac{3}{5} + \sin^{-1} \dfrac{5}{13} + \tan^{-1} \dfrac{16}{63} =$ (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The sum of the three inverse trigonometric functions simplifies to $\frac{\pi}{2}$ by converting each into an angle of a right triangle, adding two of them using the tangent addition formula, and recognizing the complementary angle relationship.

   We are asked to evaluate

   $\cos^{-1}\frac{3}{5} + \sin^{-1}\frac{5}{13} + \tan^{-1}\frac{16}{63}.$

   The key idea is to interpret each inverse trig function as an angle in a right triangle, then combine them using known identities — specifically the tangent addition formula — to see if the total is a standard angle like $\frac{\pi}{2}$, $\frac{\pi}{3}$, etc.

   ---

   1. Interpret each term as an angle in a right triangle.

      • Let $\alpha = \cos^{-1}\frac{3}{5}$. Then $\cos\alpha = \frac{3}{5}$, so in a right triangle with adjacent $3$ and hypotenuse $5$, the opposite side is $\sqrt{5^2-3^2}=4$. Hence $\tan\alpha = \frac{4}{3}$.
      • Let $\beta = \sin^{-1}\frac{5}{13}$. Then $\sin\beta = \frac{5}{13}$, so opposite $5$, hypotenuse $13$, adjacent $\sqrt{13^2-5^2}=12$. Hence $\tan\beta = \frac{5}{12}$.
      • Let $\gamma = \tan^{-1}\frac{16}{63}$. Then $\tan\gamma = \frac{16}{63}$ directly.

   2. We want $\alpha + \beta + \gamma$.

      First, combine $\alpha$ and $\beta$ using the tangent addition formula:

   $\tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3}\cdot\frac{5}{12}}.$

   Compute numerator: $\frac{4}{3} = \frac{16}{12}$, so $\frac{16}{12}+\frac{5}{12} = \frac{21}{12} = \frac{7}{4}$.

   Denominator: $1 - \frac{20}{36} = 1 - \frac{5}{9} = \frac{4}{9}$.

   Thus

   $\tan(\alpha+\beta) = \frac{7/4}{4/9} = \frac{7}{4} \cdot \frac{9}{4} = \frac{63}{16}.$

   3. Now add $\gamma$. We have $\tan(\alpha+\beta) = \frac{63}{16}$ and $\tan\gamma = \frac{16}{63}$. Notice that

   $\frac{63}{16} \cdot \frac{16}{63} = 1,$

   so $\tan(\alpha+\beta)$ and $\tan\gamma$ are reciprocals.

   For positive acute angles, if $\tan A = \frac{1}{\tan B}$, then $A + B = \frac{\pi}{2}$ (since $\tan(\frac{\pi}{2} - \theta) = \cot\theta$). …

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6. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.The number of values of $x$ satisfying $\sin 4x = \cos 3x$ and $-\dfrac{\pi}{6} < x < \dfrac{\pi}{2}$, is (A) $0$ (B) $1$ (C) $2$ (D) $3$

   ›Reveal solutionSolution

   Exactly $2$ solutions lie in the interval — option (C).

   Write $\cos 3x = \sin\!\left(\tfrac{\pi}{2}-3x\right)$, so $\sin 4x = \sin\!\left(\tfrac{\pi}{2}-3x\right)$. This gives two families:

   $4x = \frac{\pi}{2}-3x+2n\pi \implies 7x=\frac{\pi}{2}+2n\pi \implies x=\frac{\pi}{14}(1+4n),$

   $4x = \pi-\left(\frac{\pi}{2}-3x\right)+2n\pi \implies x=\frac{\pi}{2}+2n\pi.$

   Now select $x\in\left(-\tfrac{\pi}{6},\tfrac{\pi}{2}\right)\approx(-0.524,\ 1.571)$: …

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7. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $y = \sin^{-1} \left( \frac{1 - \cos 2x}{1 + \sin^4 x} \right)$ then $\frac{dy}{dx} =$ (A) $\frac{2 \cos 2x}{1 + \sin^4 x}$ (B) $\frac{2 \sin 2x}{1 + \sin^4 x}$ (C) $\frac{2 \cos 2x}{1 + \sin^8 x}$ (D) $\frac{2 \sin 2x}{1 + \sin^8 x}$

   ›Reveal solutionSolution

   The key is to simplify the argument of the inverse sine using trigonometric identities before differentiating. After simplification, the derivative becomes $\frac{2 \sin 2x}{1 + \sin^4 x}$, which matches option (B).

   ---

   We start with

   $y = \sin^{-1}\left( \frac{1 - \cos 2x}{1 + \sin^4 x} \right).$

   Concept and intuition

   The expression inside the inverse sine looks messy, but the numerator $1 - \cos 2x$ is a classic form that simplifies to $2\sin^2 x$. That’s a huge clue: the whole fraction might simplify to something like $\frac{2\sin^2 x}{1 + \sin^4 x}$, which is a perfect candidate for the identity $\sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1} t$ (for $|t|\le 1$). That substitution makes differentiation straightforward.

   ---

   Step-by-step

   1. Simplify the numerator Using the double-angle identity:

   $1 - \cos 2x = 2\sin^2 x.$

   So the argument becomes

   $\frac{2\sin^2 x}{1 + \sin^4 x}.$

   2. Recognize a standard form Let $t = \sin^2 x$. Then the argument is

   $\frac{2t}{1 + t^2}.$

   This is exactly the form $\frac{2t}{1+t^2}$, which appears in the identity

   $\sin^{-1}\left( \frac{2t}{1+t^2} \right) = 2\tan^{-1} t \quad \text{for } |t| \le 1.$

   Since $t = \sin^2 x \in [0,1]$, the condition holds. Therefore

   $y = 2\tan^{-1}(\sin^2 x).$

   3. Differentiate Differentiate $y = 2\tan^{-1}(\sin^2 x)$:

   $\frac{dy}{dx} = 2 \cdot \frac{1}{1 + (\sin^2 x)^2} \cdot \frac{d}{dx}(\sin^2 x).$

   The derivative of $\sin^2 x$ is $2\sin x \cos x = \sin 2x$. So …

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8. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $x = \sin^{-1}(\cos t)$ and $y = \tan^{-1}(\cos t)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{\cos x}{1 + \sin^2 x}$ (B) $\dfrac{\cos t}{1 + \sin^2 t}$ (C) $\dfrac{\sin t}{1 + \sin^2 t}$ (D) $\dfrac{\sin x}{1 + \cos^2 x}$

   ›Reveal solutionSolution

   The key is to express both $x$ and $y$ in terms of $t$, then use the chain rule to find $\frac{dy}{dx}$ as a function of $x$, matching option (A).

   We are given:

   $x = \sin^{-1}(\cos t), \quad y = \tan^{-1}(\cos t).$

   We need $\frac{dy}{dx}$.

   Concept and intuition:

   Both $x$ and $y$ are functions of $t$, so we can compute $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately, then use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. The trick is to simplify $\sin^{-1}(\cos t)$ into a nicer algebraic form in terms of $t$, which will make differentiation cleaner. Then we express the final answer in terms of $x$ (since the options involve $x$ or $t$).

   ---

   1. Simplify $x$ Recall: $\sin^{-1}(\cos t) = \frac{\pi}{2} - t$ for $t$ in $[0, \pi]$, but more generally we can use the identity $\sin^{-1}(\cos t) = \frac{\pi}{2} - |t|$? Actually, a cleaner approach: Let $u = \cos t$. Then $x = \sin^{-1}(u)$. But we can also write $\cos t = \sin\left(\frac{\pi}{2} - t\right)$, so

   $x = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - t\right)\right).$

   For the principal range of $\sin^{-1}$, we need $\frac{\pi}{2} - t \in [-\pi/2, \pi/2]$, i.e., $t \in [0, \pi]$. Assuming this, we get

   $x = \frac{\pi}{2} - t.$

   Hence, $t = \frac{\pi}{2} - x$.

   2. Differentiate $x$ and $y$ with respect to $t$ From $x = \frac{\pi}{2} - t$, we have

   $\frac{dx}{dt} = -1.$

   For $y = \tan^{-1}(\cos t)$, differentiate:

   $\frac{dy}{dt} = \frac{1}{1 + (\cos t)^2} \cdot (-\sin t) = -\frac{\sin t}{1 + \cos^2 t}.$

   3. Find $\frac{dy}{dx}$ Using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\frac{\sin t}{1 + \cos^2 t}}{-1} = \frac{\sin t}{1 + \cos^2 t}.$ …

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9. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.If $\theta$ is the acute angle between the curves $x^2 + y^2 = 2020\sqrt{2}$ and $x^2 - y^2 = 2020$, then

   [!FORMULA] $\frac{\sin \theta + \cos \theta}{\tan \theta} =$

   (A) $\sqrt{2}$ (B) $\frac{3 + \sqrt{3}}{2}$ (C) $\frac{3 + \sqrt{3}}{4}$ (D) $\frac{3 + \sqrt{3}}{6}$

   ›Reveal solutionSolution

   The curves cut at $45^\circ$, so $\dfrac{\sin\theta+\cos\theta}{\tan\theta}=\dfrac{\sqrt2}{1}=\sqrt2$.

   Slopes at a point of intersection.

   Circle $x^2+y^2=2020\sqrt2$: $\;2x+2y\,y'=0\Rightarrow m_1=-\dfrac{x}{y}$.

   Hyperbola $x^2-y^2=2020$: $\;2x-2y\,y'=0\Rightarrow m_2=\dfrac{x}{y}$.

   $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{-2x/y}{1-x^2/y^2}\right| =\left|\frac{-2xy}{y^2-x^2}\right|.$

   Point of intersection. Subtracting the equations, $2y^2=2020(\sqrt2-1)$ and $2x^2=2020(\sqrt2+1)$, so

   $y^2-x^2=1010\big[(\sqrt2-1)-(\sqrt2+1)\big]=-2020,$

   $x^2y^2=1010^2(\sqrt2+1)(\sqrt2-1)=1010^2\Rightarrow |xy|=1010.$

   Hence …

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10. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.If tan $15^\circ$ and tan $30^\circ$ are the roots of the equation $x^2 + px + q = 0$, then $pq =$ (A) $\dfrac{6\sqrt{3} + 10}{\sqrt{3}}$ (B) $\dfrac{10 - 6\sqrt{3}}{3}$ (C) $\dfrac{10 + 6\sqrt{3}}{3}$ (D) $\dfrac{10 - 6\sqrt{3}}{\sqrt{3}}$

    ›Reveal solutionSolution

    This problem uses Vieta's formulas to relate the roots of a quadratic equation to its coefficients. We first find the values of $\tan 15^\circ$ and $\tan 30^\circ$, then use them to determine $p$ and $q$, and finally calculate their product. The result is $\boxed{\dfrac{10 - 6\sqrt{3}}{3}}$.

    The core idea behind this problem is understanding the relationship between the roots of a polynomial equation and its coefficients. For a quadratic equation, this relationship is elegantly captured by Vieta's formulas. If we have a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, then:

    • The sum of the roots is $\alpha + \beta = -\frac{b}{a}$.
    • The product of the roots is $\alpha \beta = \frac{c}{a}$.

    In our given equation, $x^2 + px + q = 0$, the coefficient of $x^2$ is $1$. So, if the roots are $\alpha$ and $\beta$:

    • Sum of roots: $\alpha + \beta = -p$
    • Product of roots: $\alpha \beta = q$

    The problem states that the roots are $\tan 15^\circ$ and $\tan 30^\circ$. Our strategy will be to:

    1. Find the exact values of $\tan 15^\circ$ and $\tan 30^\circ$.
    2. Use these values with Vieta's formulas to find $p$ and $q$.
    3. Calculate the product $pq$.

    Let's work through the steps.

    1. Identify the roots and the equation:

       The given quadratic equation is $x^2 + px + q = 0$.

       The roots are $\alpha = \tan 15^\circ$ and $\beta = \tan 30^\circ$.

    2. Determine the values of $\tan 15^\circ$ and $\tan 30^\circ$:

       The value of $\tan 30^\circ$ is a standard trigonometric value:

       $\tan 30^\circ = \frac{1}{\sqrt{3}}$

       For $\tan 15^\circ$, we can use the tangent subtraction formula, $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. We can write $15^\circ$ as $45^\circ - 30^\circ$.

       ›Proof

       Derivation of $\tan 15^\circ$

       Let $A = 45^\circ$ and $B = 30^\circ$.

       $\tan 15^\circ = \tan(45^\circ - 30^\circ)$

       $= \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}$

       We know $\tan 45^\circ = 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

       $= \frac{1 - \frac{1}{\sqrt{3}}}{1 + (1)\left(\frac{1}{\sqrt{3}}\right)}$

       $= \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}}$

       $= \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

       To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3} - 1$:

       $= \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1}$

       $= \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2}$

       $= \frac{3 - 2\sqrt{3} + 1}{3 - 1}$

       $= \frac{4 - 2\sqrt{3}}{2}$

       $= 2 - \sqrt{3}$

       So, $\tan 15^\circ = 2 - \sqrt{3}$.

    3. Apply Vieta's formulas to find $p$ and $q$:

       The roots are $\alpha = 2 - \sqrt{3}$ and $\beta = \frac{1}{\sqrt{3}}$.

       For the equation $x^2 + px + q = 0$:

       Vieta's Formulas for $x^2 + px + q = 0$

       Sum of roots: $\alpha + \beta = -p$

       Product of roots: $\alpha \beta = q$

       • Calculate $-p$ (sum of roots): $-p = \tan 15^\circ + \tan 30^\circ$ $-p = (2 - \sqrt{3}) + \frac{1}{\sqrt{3}}$ To combine these, find a common denominator: …

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11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.If $\sinh^{-1}(-\sqrt{3}) + \cosh^{-1}(2) = K$, then $\cosh K =$ (A) $\log(2 - \sqrt{3})$ (B) $\log(2 + \sqrt{3})$ (C) $0$ (D) $1$

    ›Reveal solutionSolution

    The key idea is to evaluate the inverse hyperbolic functions as real numbers, sum them, and then compute the hyperbolic cosine of the result. The final value is 1, so the correct option is (D).

    We are given

    $\sinh^{-1}(-\sqrt{3}) + \cosh^{-1}(2) = K$

    and asked for $\cosh K$.

    ---

    Concept and Intuition

    Inverse hyperbolic functions are defined in terms of logarithms, but we can also evaluate them by recalling the definitions:

    • $\sinh^{-1} x$ is the number whose hyperbolic sine is $x$.
    • $\cosh^{-1} x$ (for $x \ge 1$) is the non-negative number whose hyperbolic cosine is $x$.

    We can compute each term exactly, then add them, and finally take $\cosh$ of the sum. A useful identity:

    $\cosh(a+b) = \cosh a \cosh b + \sinh a \sinh b$

    will let us avoid explicitly finding $K$ as a logarithm — we can directly compute $\cosh K$ from the known values of $\sinh$ and $\cosh$ of the individual terms.

    ---

    Step-by-step solution

    1. Find $\sinh^{-1}(-\sqrt{3})$ Let $a = \sinh^{-1}(-\sqrt{3})$. Then $\sinh a = -\sqrt{3}$. Recall that $\sinh a = \frac{e^a - e^{-a}}{2}$. Solving:

    $\frac{e^a - e^{-a}}{2} = -\sqrt{3} \quad\Rightarrow\quad e^a - e^{-a} = -2\sqrt{3}$

    Multiply by $e^a$:

    $e^{2a} + 2\sqrt{3}\, e^a - 1 = 0$

    Solve the quadratic in $e^a$:

    $e^a = \frac{-2\sqrt{3} \pm \sqrt{12 + 4}}{2} = \frac{-2\sqrt{3} \pm 4}{2} = -\sqrt{3} \pm 2$

    Since $e^a > 0$, we take $e^a = 2 - \sqrt{3}$ (because $2 - \sqrt{3} > 0$).

    Thus $a = \log(2 - \sqrt{3})$.

    So $\sinh^{-1}(-\sqrt{3}) = \log(2 - \sqrt{3})$.

    2. Find $\cosh^{-1}(2)$ Let $b = \cosh^{-1}(2)$. Then $\cosh b = 2$ and $b \ge 0$. Using $\cosh b = \frac{e^b + e^{-b}}{2}$:

    $\frac{e^b + e^{-b}}{2} = 2 \quad\Rightarrow\quad e^b + e^{-b} = 4$

    Multiply by $e^b$:

    $e^{2b} - 4e^b + 1 = 0$

    Solve:

    $e^b = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$ …

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12. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.If $\theta$ is an acute angle through which the coordinate axes are to be rotated about the origin in anti-clockwise direction to remove $xy$-term from the equation $4x^2 + 3xy + y^2 + 1 = 0$, then $(1 + \tan \theta)^2 =$ (A) $2$ (B) $4$ (C) $1$ (D) $3$

    ›Reveal solutionSolution

    To eliminate the $xy$-term in a rotated conic, we use $\cot 2\theta = \frac{A-C}{B}$. Here $A=4$, $B=3$, $C=1$, so $\cot 2\theta = 1$, giving $\tan\theta = \sqrt{2}-1$ and $(1+\tan\theta)^2 = 2$. The correct option is (A).

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    Concept & Intuition

    When we rotate the coordinate axes by an angle $\theta$, the $xy$-term in a general quadratic $Ax^2 + Bxy + Cy^2 + \dots = 0$ disappears if $\theta$ satisfies

    $\cot 2\theta = \frac{A - C}{B}.$

    This formula comes from the transformation of the quadratic form under rotation: the coefficient of $x'y'$ becomes $B\cos 2\theta - (A-C)\sin 2\theta$, and setting it to zero yields the condition above. For an acute $\theta$, we take the positive root for $\tan\theta$ from the double-angle identity.

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    Step-by-step solution

    1. Identify coefficients

       The given equation is $4x^2 + 3xy + y^2 + 1 = 0$.

       Here $A = 4$, $B = 3$, $C = 1$. The constant term $1$ does not affect the rotation condition.

    2. Apply the rotation condition

       To remove the $xy$-term, we need

    $\cot 2\theta = \frac{A - C}{B} = \frac{4 - 1}{3} = \frac{3}{3} = 1.$

    So $\cot 2\theta = 1$, meaning $2\theta = 45^\circ$ (since $\theta$ is acute, $2\theta$ is acute as well). Thus $\theta = 22.5^\circ$.

    3. Find $\tan\theta$ We know $\cot 2\theta = 1 \implies \tan 2\theta = 1$. Use the double-angle identity:

    $\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} = 1.$

    Let $t = \tan\theta$. Then

    $$$$

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