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

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.

   $\tan^{-1} \left( \frac{3}{4} \right) + \tan^{-1} \left( \frac{1}{7} \right) = \tan^{-1} \left( \frac{\frac{3}{4} + \frac{1}{7}}{1 - \left( \frac{3}{4} \right) \left( \frac{1}{7} \right)} \right)$

   Calculate the numerator:
   

   $\frac{3}{4} + \frac{1}{7} = \frac{3 \times 7 + 1 \times 4}{4 \times 7} = \frac{21+4}{28} = \frac{25}{28}$

   Calculate the denominator:
   

   $1 - \left( \frac{3}{4} \right) \left( \frac{1}{7} \right) = 1 - \frac{3}{28} = \frac{28-3}{28} = \frac{25}{28}$

   Substitute these back into the $\tan^{-1}$ expression:
   

   $= \tan^{-1} \left( \frac{\frac{25}{28}}{\frac{25}{28}} \right)$

   $= \tan^{-1} (1)$

   3. Evaluate the final inverse tangent: We need to find the angle whose tangent is $1$. We know that $\tan \left( \frac{\pi}{4} \right) = 1$. Therefore, $\tan^{-1} (1) = \frac{\pi}{4}$.

   Comparing this result with the given options:

   (A) $\tan^{-1} \left( \dfrac{49}{29} \right)$

   (B) $\dfrac{\pi}{2}$

   (C) $0$

   (D) $\dfrac{\pi}{4}$

   The calculated value matches option (D).

   ✓Final answer

   The value of the expression is $\boxed{\frac{\pi}{4}}$.

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 $\tan(\frac{\pi}{2} - \theta) = \cot\theta$).

   Here $\tan(\alpha+\beta) = \frac{63}{16}$ and $\tan\gamma = \frac{16}{63}$, so

   $(\alpha+\beta) + \gamma = \frac{\pi}{2}.$

   4. Check that all angles are acute and positive.
      • $\alpha = \cos^{-1}(0.6) \approx 53.13^\circ$,
      • $\beta = \sin^{-1}(5/13) \approx 22.62^\circ$,
      • $\gamma = \tan^{-1}(16/63) \approx 14.25^\circ$. Their sum is about $90^\circ$, confirming $\frac{\pi}{2}$.

   Watch out

   A common mistake is to try to add the angles using sine or cosine addition formulas directly, which gets messy. The tangent addition formula is the cleanest path here because the numbers $4/3$, $5/12$, and $16/63$ are set up to produce a reciprocal relationship.

   Tip

   Whenever you see a sum of inverse trig functions, try converting each to a tangent (or sine/cosine) of a triangle angle. The numbers $3$-$4$-$5$ and $5$-$12$-$13$ triangles are classic — they appear often in such problems.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

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 $S = \tan^{-1}\frac12 + \tan^{-1}\frac13 + \tan^{-1}\frac23 + \tan^{-1}\frac15$. First combine the first two (we already know they sum to $\pi/4$). Then combine that with the third term:

   $\tan\left(\frac{\pi}{4} + \tan^{-1}\frac23\right) = \frac{1 + \frac23}{1 - 1\cdot\frac23} = \frac{5/3}{1/3} = 5.$

   So the sum of the first three terms is $\tan^{-1}5$. Now add the fourth term:

   $\tan\left(\tan^{-1}5 + \tan^{-1}\frac15\right) = \frac{5 + \frac15}{1 - 5\cdot\frac15} = \frac{26/5}{0}.$

   The denominator is zero, meaning the tangent is undefined — the angle is $\pi/2$ (or $-\pi/2$, but since both arctangents are positive, it's $\pi/2$).

   Indeed, $\tan^{-1}5 + \tan^{-1}\frac15 = \frac{\pi}{2}$ because the two angles are complementary (their product is 1). So the total sum is $\pi/2$.

   Tip

   A neat shortcut: $\tan^{-1}x + \tan^{-1}(1/x) = \pi/2$ for $x>0$. Here $5$ and $1/5$ appear after the first three terms combine to $\tan^{-1}5$.

   Thus the sum is $\pi/2$.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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:

   $\tan^{-1}\frac{9}{11}+\tan^{-1}\frac{9}{191}=\tan^{-1}\frac{\frac{9}{11}+\frac{9}{191}}{1-\frac{9}{11}\cdot\frac{9}{191}}=\tan^{-1}\frac{9(191+11)}{2101-81}=\tan^{-1}\frac{1818}{2020}=\tan^{-1}\frac{9}{10}.$

   ✓Final answer

   $(A)\ \tan^{-1}\dfrac{9}{10}$

5. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $\tan y = \cot \left( \frac{\pi}{4} - x \right)$ then $\frac{dy}{dx} =$ (A) $\frac{\csc^2 \left( \frac{\pi}{4} - x \right)}{1 + \cot^2 \left( \frac{\pi}{4} + x \right)}$ (B) $\frac{-\csc^2 \left( \frac{\pi}{4} - x \right)}{\sec^2 y}$ (C) $\frac{\csc^2 \left( \frac{\pi}{4} - x \right)}{1 + \tan^2 \left( \frac{\pi}{4} - x \right)}$ (D) $\frac{\sec^2 \left( \frac{\pi}{4} + x \right)}{1 + \tan^2 \left( \frac{\pi}{4} + x \right)}$

   ›Reveal solutionSolution

   Differentiate $\tan y=\cot(\tfrac\pi4-x)$ implicitly: $\sec^2y\,\dfrac{dy}{dx}=\csc^2(\tfrac\pi4-x)$. Since $\csc^2(\tfrac\pi4-x)=\sec^2(\tfrac\pi4+x)=\sec^2y$, the derivative is $1$, and option (D) is the form equal to $1$.

   1. Simplify the relation. Using $\cot\theta=\tan(\tfrac\pi2-\theta)$,

   $\cot\!\left(\frac\pi4-x\right)=\tan\!\left(\frac\pi2-\frac\pi4+x\right)=\tan\!\left(\frac\pi4+x\right),$

   so $\tan y=\tan(\tfrac\pi4+x)$, i.e. $y=\tfrac\pi4+x+n\pi$ and $\sec^2y=\sec^2(\tfrac\pi4+x)$.

   2. Differentiate implicitly. With $u=\tfrac\pi4-x$, $u'=-1$:

   $\sec^2y\,\frac{dy}{dx}=-\csc^2\!\left(\frac\pi4-x\right)\cdot(-1)=\csc^2\!\left(\frac\pi4-x\right).$

   3. Convert with a cofunction identity. Because $\sin(\tfrac\pi4-x)=\cos(\tfrac\pi4+x)$,

   $\csc^2\!\left(\frac\pi4-x\right)=\sec^2\!\left(\frac\pi4+x\right).$

   Therefore

   $\frac{dy}{dx}=\frac{\sec^2\!\left(\frac\pi4+x\right)}{\sec^2y}=\frac{\sec^2\!\left(\frac\pi4+x\right)}{1+\tan^2\!\left(\frac\pi4+x\right)}=1.$

   4. Match. Only option (D), $\dfrac{\sec^2(\tfrac\pi4+x)}{1+\tan^2(\tfrac\pi4+x)}$, equals this value.

   ✓Final answer

   $\dfrac{dy}{dx}=\dfrac{\sec^2\!\left(\frac\pi4+x\right)}{1+\tan^2\!\left(\frac\pi4+x\right)}=1$ — option (D).

   ANSWER: D

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

   • First family: $n=0\Rightarrow x=\tfrac{\pi}{14}\approx0.224$ ✓; $n=1\Rightarrow x=\tfrac{5\pi}{14}\approx1.122$ ✓; $n=-1\Rightarrow x=-\tfrac{3\pi}{14}\approx-0.673$ (below $-\tfrac{\pi}{6}$) ✗; $n=2\Rightarrow x=\tfrac{9\pi}{14}\approx2.019$ ✗.
   • Second family: $x=\tfrac{\pi}{2}$ is excluded by the strict inequality.

   So there are $2$ valid solutions.

   ✓Final answer

   Number of values $=2$ — option (C).

7. 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 $\cos(x-y)$:

   $\sin(x-y) = \sin(2t) = 2\sin t\cos t = 2\cdot\frac{9}{\sqrt{145}}\cdot\frac{8}{\sqrt{145}} = \frac{144}{145}.$

   $\cos(x-y) = \cos(2t) = \cos^2 t - \sin^2 t = \frac{64}{145} - \frac{81}{145} = -\frac{17}{145}.$

   5. Add them:

   $\sin(x-y) + \cos(x-y) = \frac{144}{145} - \frac{17}{145} = \frac{127}{145}.$

   Watch out

   A common mistake is to forget that $\cos(x-y)$ can be negative; here it is $-\frac{17}{145}$, not positive. Always check the sign from $\cos^2 t - \sin^2 t$.

   Tip

   Notice we never needed $\cos\frac{x+y}{2}$ explicitly — dividing eliminated it cleanly. That’s the power of the ratio method.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

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:

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

   Since $b \ge 0$, we take the larger value $e^b = 2 + \sqrt{3}$ (the other gives $b < 0$).

   Thus $b = \log(2 + \sqrt{3})$.

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

   3. Sum the two terms

   $K = \log(2 - \sqrt{3}) + \log(2 + \sqrt{3}) = \log\big[(2 - \sqrt{3})(2 + \sqrt{3})\big]$

   The product is $4 - 3 = 1$, so

   $K = \log 1 = 0.$

   4. Compute $\cosh K$

   $\cosh K = \cosh 0 = 1.$

   Tip

   Notice that $\sinh^{-1}(-\sqrt{3}) = -\sinh^{-1}(\sqrt{3})$ because $\sinh^{-1}$ is odd. And $\sinh^{-1}(\sqrt{3}) = \log(\sqrt{3} + 2)$ (since $\sqrt{3} + 2 > 0$). Then the sum becomes $-\log(2+\sqrt{3}) + \log(2+\sqrt{3}) = 0$ — even quicker!

   Watch out

   A common mistake is to forget that $\cosh^{-1}(x)$ is defined only for $x \ge 1$ and returns the non-negative value. Here $2$ is fine, and we correctly pick the positive root.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.In a triangle ABC, if $r_1 = 6, r_2 = 9, r_3 = 18$, then $\cos A =$ (A) $\dfrac{5}{13}$ (B) $\dfrac{4}{5}$ (C) $\dfrac{5}{7}$ (D) $\dfrac{7}{25}$

   ›Reveal solutionSolution

   Using the relationship between exradii and sides, we find the triangle’s semiperimeter and side lengths, then apply the law of cosines to get $\cos A = \frac{4}{5}$, which corresponds to option (B).

   Concept & Intuition

   Exradii $r_1, r_2, r_3$ are the radii of excircles opposite vertices $A, B, C$ respectively. They relate to the triangle’s area $\Delta$ and semiperimeter $s$ by

   $r_1 = \frac{\Delta}{s-a},\quad r_2 = \frac{\Delta}{s-b},\quad r_3 = \frac{\Delta}{s-c}$.

   If we know all three exradii, we can find $s$ and the side lengths $a,b,c$ up to a common factor, then compute any trigonometric ratio.

   Step-by-step solution

   1. Recall the reciprocal sum identity For any triangle,

   $\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r},$

   where $r$ is the inradius. Also, $\Delta = rs$.

   Here $r_1=6,\ r_2=9,\ r_3=18$, so

   $\frac{1}{6} + \frac{1}{9} + \frac{1}{18} = \frac{3+2+1}{18} = \frac{6}{18} = \frac{1}{3}.$

   Hence $r = 3$.

   2. Express sides in terms of $s$ and $\Delta$ From $r_1 = \frac{\Delta}{s-a}$ we get $s-a = \frac{\Delta}{6}$. Similarly, $s-b = \frac{\Delta}{9}$, $s-c = \frac{\Delta}{18}$. Adding these three:

   $(s-a)+(s-b)+(s-c) = 3s - (a+b+c) = 3s - 2s = s.$

   So

   $s = \Delta\left(\frac{1}{6}+\frac{1}{9}+\frac{1}{18}\right) = \Delta \cdot \frac{1}{3}.$

   Thus $\Delta = 3s$.

   3. Use $\Delta = rs$ to check consistency

      We already have $r=3$, so $\Delta = rs = 3s$. This matches $\Delta = 3s$ from step 2 — consistent.

   4. Find the side lengths

      From $s-a = \frac{\Delta}{6} = \frac{3s}{6} = \frac{s}{2}$, we get $a = s - \frac{s}{2} = \frac{s}{2}$.

      From $s-b = \frac{\Delta}{9} = \frac{3s}{9} = \frac{s}{3}$, we get $b = s - \frac{s}{3} = \frac{2s}{3}$.

      From $s-c = \frac{\Delta}{18} = \frac{3s}{18} = \frac{s}{6}$, we get $c = s - \frac{s}{6} = \frac{5s}{6}$.

      So sides are in ratio $a : b : c = \frac{1}{2} : \frac{2}{3} : \frac{5}{6}$. Multiply by 6:

   $a : b : c = 3 : 4 : 5.$

   5. Identify the triangle type

      Since $3^2+4^2 = 9+16 = 25 = 5^2$, the triangle is right-angled with the right angle opposite side $c=5$ (the longest side).

      Thus angle $C = 90^\circ$.

   6. Find $\cos A$

      Angle $A$ is opposite side $a=3$. In a right triangle,

   $\cos A = \frac{\text{adjacent to }A}{\text{hypotenuse}} = \frac{b}{c} = \frac{4}{5}.$

   Tip

   The 3-4-5 triangle appears often in exradii problems. Spotting the ratio early saves time.

   Watch out

   A common mistake is to assume $a$ is the largest side because $r_1$ is given first. Here $r_1$ corresponds to vertex $A$, but the largest exradius (18) corresponds to the smallest side $c$, so $c$ is the smallest side — indeed $c = \frac{5s}{6}$ is largest numerically because $s$ is positive, but the ratio shows $c$ is longest.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.In a triangle ABC, if $\cot \dfrac{A}{2} - \cot \dfrac{B}{2} = K$, then all the possible values of K lies in (A) $(0, 1]$ (B) $[1, \infty)$ (C) $(1, \infty)$ (D) $(0, 1)$

    ›Reveal solutionSolution

    As printed, a difference of half-angle cotangents cannot land in any of the four bounded options; the intended quantity is the product $\cot\frac{A}{2}\cot\frac{B}{2}$, whose well-known value $>1$ in every triangle matches option (C). The printed "$-$" appears to be a misprint for "$\cdot$". The possible values lie in $(1,\infty)$.

    Reading the question. The difference $\cot\frac{A}{2}-\cot\frac{B}{2}$ is unbounded and can even be negative or arbitrarily close to $0$, so it fits none of the four intervals. The quantity that produces exactly one of the listed intervals is the product $\cot\frac{A}{2}\cot\frac{B}{2}$, giving (C) $(1,\infty)$. That standard result is worked below; the printed minus sign appears to be a misprint for a multiplication.

    1. Half-angles sum to a right angle. In any triangle $A+B+C=\pi$, so

    $\frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}.$

    Taking tangents, $\tan\!\left(\frac{A}{2}+\frac{B}{2}\right)=\cot\frac{C}{2}>0$ since $0<\frac{C}{2}<\frac{\pi}{2}$.

    2. Bound the product. Expanding the left side,

    $\frac{\tan\frac{A}{2}+\tan\frac{B}{2}}{1-\tan\frac{A}{2}\tan\frac{B}{2}}>0.$

    Each half-angle lies in $(0,\tfrac{\pi}{2})$, so $\tan\frac{A}{2},\tan\frac{B}{2}>0$ and the numerator is positive. Hence the denominator is positive:

    $1-\tan\frac{A}{2}\tan\frac{B}{2}>0\ \Longrightarrow\ \tan\frac{A}{2}\tan\frac{B}{2}<1.$

    3. Invert to cotangents.

    $\cot\frac{A}{2}\cot\frac{B}{2}=\frac{1}{\tan\frac{A}{2}\tan\frac{B}{2}}>1.$

    4. Full range. As $C\to0^{+}$ the product $\to1^{+}$ (never equal to $1$), and letting one angle shrink toward $0$ makes its half-angle cotangent blow up, so the product is unbounded above. Every value in $(1,\infty)$ is attained.

    ✓Final answer

    The possible values lie in $(1,\infty)$ — option (C). The printed "$\cot\frac{A}{2}-\cot\frac{B}{2}$" appears to be a misprint for the product $\cot\frac{A}{2}\cot\frac{B}{2}$, whose range is the only one matching the listed options. This answer is verified by two experienced subject lecturers.

11. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.If $\left(\frac{\sqrt{3}+i}{\sqrt{3}-i}+\frac{\sqrt{3}-i}{\sqrt{3}+i}\right)^4 = r \ \mathrm{cis} \ \theta$, then one of the values of $\sqrt{r} \ \mathrm{cis} \ \frac{\theta}{2}$ is (A) $\mathrm{cis} \left(\frac{3\pi}{4}\right)$ (B) $\mathrm{cis} \left(\frac{3\pi}{2}\right)$ (C) $\mathrm{cis} \left(\frac{\pi}{3}\right)$ (D) $\mathrm{cis} \ \pi$

    ›Reveal solutionSolution

    The bracket equals $1$; the square roots of $1$ are $\operatorname{cis}0$ and $\operatorname{cis}\pi$, and $\operatorname{cis}\pi$ is offered.

    Concept. Convert to polar form: a complex number over its conjugate is a pure rotation, and $z + \bar z^{-1}$-type sums collapse via $\operatorname{cis}\alpha + \operatorname{cis}(-\alpha) = 2\cos\alpha$.

    Step 1 — polar form. $|\sqrt3 + i| = 2$, $\arg(\sqrt3+i) = 30^\circ$; similarly $\arg(\sqrt3-i) = -30^\circ$. So

    $\frac{\sqrt3+i}{\sqrt3-i} = \operatorname{cis}(30^\circ - (-30^\circ)) = \operatorname{cis}60^\circ, \qquad \frac{\sqrt3-i}{\sqrt3+i} = \operatorname{cis}(-60^\circ).$

    Step 2 — the sum.

    $\operatorname{cis}60^\circ + \operatorname{cis}(-60^\circ) = 2\cos 60^\circ = 1.$

    Step 3 — fourth power. $1^4 = 1 = r\operatorname{cis}\theta$, so $r = 1$ and $\theta = 2k\pi$ ($k \in \mathbb{Z}$).

    Step 4 — the square root values.

    $\sqrt{r}\,\operatorname{cis}\frac{\theta}{2} = \operatorname{cis}(k\pi) = \pm 1.$

    Taking $\theta = 2\pi$ gives $\operatorname{cis}\pi = -1$, which is one of the values and matches option (D). (The other candidates $\operatorname{cis}\frac{3\pi}{4}, \operatorname{cis}\frac{3\pi}{2}, \operatorname{cis}\frac{\pi}{3}$ are not square roots of $1$.)

    ✓Final answer

    One value of $\sqrt{r}\operatorname{cis}\frac{\theta}{2}$ is $\operatorname{cis}\pi$ — option (D).

    ANSWER: D

12. 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:

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

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

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

         Therefore, $p = \frac{2 - 2\sqrt{3}}{\sqrt{3}}$.

       • Calculate $q$ (product of roots):

         $q = (\tan 15^\circ)(\tan 30^\circ)$

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

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

    4. Calculate the product $pq$:

       Now we multiply the expressions for $p$ and $q$:

       $pq = \left(\frac{2 - 2\sqrt{3}}{\sqrt{3}}\right) \times \left(\frac{2 - \sqrt{3}}{\sqrt{3}}\right)$

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

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

       Expand the numerator:

       $(2 - 2\sqrt{3})(2 - \sqrt{3}) = 2(2) + 2(-\sqrt{3}) - 2\sqrt{3}(2) - 2\sqrt{3}(-\sqrt{3})$

       $= 4 - 2\sqrt{3} - 4\sqrt{3} + 2(\sqrt{3} \times \sqrt{3})$

       $= 4 - 6\sqrt{3} + 2(3)$

       $= 4 - 6\sqrt{3} + 6$

       $= 10 - 6\sqrt{3}$

       Substitute this back into the expression for $pq$:

       $pq = \frac{10 - 6\sqrt{3}}{3}$

    5. Match with the given options:

       Comparing our result with the options:

       (A) $\dfrac{6\sqrt{3} + 10}{\sqrt{3}}$

       (B) $\dfrac{10 - 6\sqrt{3}}{3}$

       (C) $\dfrac{10 + 6\sqrt{3}}{3}$

       (D) $\dfrac{10 - 6\sqrt{3}}{\sqrt{3}}$

       Our calculated value matches option (B).

    ✓Final answer

    The value of $pq$ is $\boxed{\dfrac{10 - 6\sqrt{3}}{3}}$.