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 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 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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3. 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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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 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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6. 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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7. 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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8. 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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9. 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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10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.In $\triangle ABC$, if $\dfrac{\cos A}{a} = \dfrac{\cos B}{b} = \dfrac{\cos C}{c}$ and side $a = 2$, then area of the $\triangle ABC$ (in sq. units) is (A) $8\sqrt{2}$ (B) $4\sqrt{3}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\sqrt{3}$

    ›Reveal solutionSolution

    The given condition, combined with the Sine Rule, implies that the triangle is equilateral. With side $a=2$, the area is $\boxed{\sqrt{3}}$.

    The problem provides a relationship between the cosines of the angles and the lengths of the sides of a triangle, and asks for its area given one side length. The key to solving this is to use the Sine Rule to express the side lengths in terms of the sines of the angles. This substitution will simplify the given condition, allowing us to determine the type of triangle. Once the type of triangle is known, calculating its area becomes straightforward.

    Here's how we can approach this:

    1. Recall the Sine Rule: For any triangle $\triangle ABC$ with sides $a, b, c$ opposite to angles $A, B, C$ respectively, the Sine Rule states:

    $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

    where $k$ is a constant. This means we can write $a = k \sin A$, $b = k \sin B$, and $c = k \sin C$.
    

    2. Substitute into the given condition: We are given the condition $\dfrac{\cos A}{a} = \dfrac{\cos B}{b} = \dfrac{\cos C}{c}$.

    Substitute the expressions for $a, b, c$ from the Sine Rule into this condition:

    $\frac{\cos A}{k \sin A} = \frac{\cos B}{k \sin B} = \frac{\cos C}{k \sin C}$

    3. Simplify the expression: Since $k$ is a non-zero constant (as side lengths are non-zero), we can cancel $k$ from the denominator of each term:

    $\frac{\cos A}{\sin A} = \frac{\cos B}{\sin B} = \frac{\cos C}{\sin C}$

    Recognizing that $\frac{\cos x}{\sin x} = \cot x$, this simplifies to:
    

    $\cot A = \cot B = \cot C$

    4. Determine the type of triangle: For angles $A, B, C$ in a triangle, which must be between $0^\circ$ and $180^\circ$ (i.e., $0 < A, B, C < \pi$), if their cotangents are equal, then the angles themselves must be equal. …

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11. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
    Q.$\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} =$ (A) $\frac{2}{3}$ (B) $-\frac{1}{6}$ (C) $-\frac{1}{12}$ (D) $\frac{1}{2}$

    ›Reveal solutionSolution

    The limit simplifies by factoring out a common power of $\cos x$ and using the series expansions for $\cos x$ and $(1+u)^\alpha$; the final value is $-\frac{1}{12}$, which corresponds to option (C).

    We want

    $L = \lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}.$

    Both numerator and denominator vanish as $x \to 0$, so this is a $\frac{0}{0}$ form. The key is to rewrite the numerator in terms of a common factor and then use expansions near $x=0$.

    ---

    1. Factor out the smallest power of $\cos x$.

    Write $\sqrt{\cos x} = (\cos x)^{1/2}$ and $\sqrt[3]{\cos x} = (\cos x)^{1/3}$. The smaller exponent is $\frac{1}{3}$, so factor $(\cos x)^{1/3}$ out:

    $(\cos x)^{1/2} - (\cos x)^{1/3} = (\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right].$

    Thus

    $L = \lim_{x \to 0} \frac{(\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right]}{\sin^2 x}.$

    As $x \to 0$, $\cos x \to 1$, so $(\cos x)^{1/3} \to 1$. That factor is harmless; the interesting part is the bracket.

    ---

    2. Expand $\cos x$ near $0$.

    We know

    $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6).$

    Also $\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$.

    ---

    3. Expand $(\cos x)^{1/6}$ using $(1+u)^\alpha$.

    Let $u = \cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} + \cdots$. Then

    $(\cos x)^{1/6} = (1+u)^{1/6} = 1 + \frac{1}{6}u + \frac{(1/6)(1/6 - 1)}{2}u^2 + \cdots$

    $= 1 + \frac{1}{6}\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + \frac{1}{6}\left(-\frac{5}{6}\right)\frac{1}{2}\left(-\frac{x^2}{2}\right)^2 + O(x^6).$

    Compute term by term:

    • Linear in $u$: $\frac{1}{6}\left(-\frac{x^2}{2}\right) = -\frac{x^2}{12}$, and the $\frac{x^4}{24}$ part gives $+\frac{x^4}{144}$.
    • Quadratic in $u$: $\frac{1}{6}\cdot\left(-\frac{5}{6}\right)\cdot\frac{1}{2} = -\frac{5}{72}$, times $u^2 = \left(-\frac{x^2}{2}\right)^2 = \frac{x^4}{4}$, gives $-\frac{5}{72}\cdot\frac{x^4}{4} = -\frac{5x^4}{288}$.

    So

    $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} + \left(\frac{1}{144} - \frac{5}{288}\right)x^4 + O(x^6).$

    The $x^4$ coefficient: $\frac{1}{144} = \frac{2}{288}$, so $\frac{2}{288} - \frac{5}{288} = -\frac{3}{288} = -\frac{1}{96}$.

    Thus

    $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} - \frac{x^4}{96} + O(x^6).$

    ---

    4. Assemble the limit. …

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12. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.$\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} =$ (A) $\frac{1}{2}$ (B) $-\frac{1}{12}$ (C) $-\frac{1}{6}$ (D) $\frac{2}{3}$

    ›Reveal solutionSolution

    The limit simplifies by factoring out $\sqrt[3]{\cos x}$ and using series expansions for $\cos x$ and the binomial expansion; the result is $-\frac{1}{12}$, so the correct option is (B).

    Concept & Intuition

    When $x \to 0$, both numerator and denominator approach $0$, giving a $\frac{0}{0}$ form. Direct substitution fails. The trick: rewrite the numerator as a common power of $\cos x$, then expand $\cos x$ as a series near $0$: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots$. The square and cube roots become binomial expansions $(1+u)^p \approx 1 + pu + \frac{p(p-1)}{2}u^2 + \cdots$, which lets us isolate the leading-order cancellation. The denominator $\sin^2 x \approx x^2$ sets the scale.

    Step-by-step solution

    1. Rewrite the numerator with a common factor Let $t = \cos x$. Then the numerator is $t^{1/2} - t^{1/3}$. Factor out $t^{1/3}$:

    $\sqrt{\cos x} - \sqrt[3]{\cos x} = (\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right].$

    As $x \to 0$, $\cos x \to 1$, so $(\cos x)^{1/3} \to 1$ and won't affect the limit's value. The key is the bracket $(\cos x)^{1/6} - 1$.

    2. Expand $\cos x$ near $0$

    $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6).$

    Let $u = -\frac{x^2}{2} + \frac{x^4}{24} + \cdots$, so $\cos x = 1 + u$ with $u \to 0$.

    3. Binomial expansion for $(1+u)^{1/6}$

    $(1+u)^{1/6} = 1 + \frac{1}{6}u + \frac{(1/6)(1/6 - 1)}{2}u^2 + O(u^3).$

    Compute $u$ and $u^2$ to order $x^4$:

    • $u = -\frac{x^2}{2} + \frac{x^4}{24} + \cdots$
    • $u^2 = \left(-\frac{x^2}{2}\right)^2 + \cdots = \frac{x^4}{4} + \cdots$ (higher terms are $O(x^6)$).

    Then

    $(1+u)^{1/6} = 1 + \frac{1}{6}\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + \frac{(1/6)(-5/6)}{2}\cdot\frac{x^4}{4} + O(x^6).$

    Simplify:

    • Linear term: $-\frac{x^2}{12} + \frac{x^4}{144}$
    • Quadratic term: $\frac{-5/36}{2} \cdot \frac{x^4}{4} = -\frac{5}{72} \cdot \frac{x^4}{4} = -\frac{5x^4}{288}$.

    So

    $(\cos x)^{1/6} = 1 - \frac{x^2}{12} + \left(\frac{1}{144} - \frac{5}{288}\right)x^4 + O(x^6) = 1 - \frac{x^2}{12} - \frac{3}{288}x^4 + \cdots = 1 - \frac{x^2}{12} - \frac{x^4}{96} + \cdots.$

    4. Thus the bracket

    $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} - \frac{x^4}{96} + O(x^6).$

    5. Denominator expansion $$\sin^2 x = \left(x - \frac{x^3}{6} + \cdots\right)^2 = x^2 - \frac{x^4}{3} + O(x^6). …$$

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