Q.If $a_1, a_2, a_3, \ldots, a_n$ is an arithmetic progression with common difference $d$, then evaluate the following expression: $\tan\left[\tan^{-1}\frac{d}{1+a_1a_2}+\tan^{-1}\frac{d}{1+a_2a_3}+\tan^{-1}\frac{d}{1+a_3a_4}+\cdots+\tan^{-1}\frac{d}{1+a_{n-1}a_n}\right]$.

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.If $\tan^{-1}\left[\dfrac{1}{1+1.2}\right] + \tan^{-1}\left[\dfrac{1}{1+2.3}\right] + \cdots + \tan^{-1}\left[\dfrac{1}{1+n(n+1)}\right] = \tan^{-1}[x]$, then $x =$ (A) $\dfrac{1}{n+1}$ (B) $\dfrac{n}{n+1}$ (C) $\dfrac{1}{n+2}$ (D) $\dfrac{n}{n+2}$

   ›Reveal solutionSolution

   Each term telescopes as $\tan^{-1}(k+1)-\tan^{-1}(k)$; summing collapses the whole series to $\tan^{-1}\big(\frac{n}{n+2}\big)$, so $x=\dfrac{n}{n+2}$.

   Concept and Intuition

   The key identity is $\tan^{-1}\dfrac{1}{1+k(k+1)} = \tan^{-1}(k+1)-\tan^{-1}k$, which follows from the tangent subtraction formula $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\dfrac{p-q}{1+pq}$ applied with $p=k+1,\,q=k$. Once each term is in this "difference" form, the whole sum telescopes, leaving only the first and last pieces.

   Step-by-Step Solution

   1. Verify the telescoping identity: $\tan^{-1}(k+1)-\tan^{-1}(k) = \tan^{-1}\dfrac{(k+1)-k}{1+(k+1)k} = \tan^{-1}\dfrac{1}{1+k(k+1)}$. ✓ matches each term's form (with $k=1,2,\dots,n$).
   2. So the sum $\sum_{k=1}^{n}\tan^{-1}\dfrac{1}{1+k(k+1)} = \sum_{k=1}^{n}\left[\tan^{-1}(k+1)-\tan^{-1}(k)\right]$.
   3. This telescopes: all intermediate terms cancel, leaving $\tan^{-1}(n+1) - \tan^{-1}(1)$.
   4. Apply the subtraction formula again: $\tan^{-1}(n+1)-\tan^{-1}(1) = \tan^{-1}\dfrac{(n+1)-1}{1+(n+1)(1)} = \tan^{-1}\dfrac{n}{n+2}$. …

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2. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\cot\left(\displaystyle\sum_{n=1}^{50}\operatorname{Tan}^{-1}\left(\dfrac{1}{1+n+n^2}\right)\right)=$ (A) $\dfrac{26}{25}$ (B) $\dfrac{25}{26}$ (C) $\dfrac{50}{51}$ (D) $\dfrac{52}{51}$

   ›Reveal solutionSolution

   This is a telescoping sum of inverse tangents; the sum collapses to $\tan^{-1}51-\tan^{-1}1$, and its cotangent is $26/25$.

   Concept and Intuition

   The identity $\tan^{-1}(n+1)-\tan^{-1}(n)=\tan^{-1}\!\left(\dfrac{1}{1+n(n+1)}\right)$ (from the tangent subtraction formula) turns each term of the sum into a telescoping difference, so almost every intermediate value cancels, leaving only the first and last terms.

   Step-by-Step Solution

   1. Note $1+n+n^2 = 1+n(n+1)$, matching the tangent-subtraction denominator form $1+ab$ with $a=n+1, b=n$.
   2. So $\tan^{-1}\!\left(\dfrac1{1+n+n^2}\right) = \tan^{-1}(n+1)-\tan^{-1}(n)$.
   3. Summing from $n=1$ to $50$ telescopes: $\sum_{n=1}^{50}\left[\tan^{-1}(n+1)-\tan^{-1}(n)\right] = \tan^{-1}(51)-\tan^{-1}(1)$.
   4. Let $\theta=\tan^{-1}(51)$, so $\tan\theta=51$. We need $\cot\left(\theta-\dfrac\pi4\right)$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.$\operatorname{Tan}^{-1}\left(\dfrac{1}{2}\right) + \operatorname{Tan}^{-1}\left(\dfrac{1}{8}\right) + \operatorname{Tan}^{-1}\left(\dfrac{1}{18}\right) + \operatorname{Tan}^{-1}\left(\dfrac{1}{32}\right) =$ (A) $\operatorname{Tan}^{-1}\left(\dfrac{3}{5}\right)$ (B) $\operatorname{Tan}^{-1}\left(\dfrac{5}{8}\right)$ (C) $\operatorname{Tan}^{-1}\left(\dfrac{3}{4}\right)$ (D) $\operatorname{Tan}^{-1}\left(\dfrac{4}{5}\right)$

   ›Reveal solutionSolution

   Recognize the telescoping identity $\tan^{-1}\frac{1}{2n^2}=\tan^{-1}\frac{1}{2n-1}-\tan^{-1}\frac{1}{2n+1}$; the four given terms ($n=1,2,3,4$) telescope down to $\tan^{-1}1-\tan^{-1}\frac19=\tan^{-1}\frac45$.

   Concept and Intuition

   The denominators $2,8,18,32$ are exactly $2\cdot1^2,\ 2\cdot2^2,\ 2\cdot3^2,\ 2\cdot4^2$ — a strong hint to use the identity $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\dfrac{a-b}{1+ab}$ in reverse: for consecutive odd-reciprocal terms $\frac{1}{2n-1}$ and $\frac{1}{2n+1}$, their difference is exactly $\tan^{-1}\frac{1}{2n^2}$. This turns the whole sum into a telescoping series where all the intermediate terms cancel.

   Step-by-Step Solution

   1. Verify the identity for general $n$: $\dfrac{\frac{1}{2n-1}-\frac{1}{2n+1}}{1+\frac{1}{(2n-1)(2n+1)}} = \dfrac{\frac{2}{4n^2-1}}{\frac{4n^2-1+1}{4n^2-1}} = \dfrac{2}{4n^2}=\dfrac{1}{2n^2}$. So $\tan^{-1}\frac{1}{2n^2}=\tan^{-1}\frac{1}{2n-1}-\tan^{-1}\frac{1}{2n+1}$.
   2. Apply with $n=1,2,3,4$:

   • $\tan^{-1}\frac12 = \tan^{-1}1-\tan^{-1}\frac13$
   • $\tan^{-1}\frac18 = \tan^{-1}\frac13-\tan^{-1}\frac15$
   • $\tan^{-1}\frac{1}{18} = \tan^{-1}\frac15-\tan^{-1}\frac17$
   • $\tan^{-1}\frac{1}{32} = \tan^{-1}\frac17-\tan^{-1}\frac19$

   3. Summing all four, the intermediate terms $\tan^{-1}\frac13,\tan^{-1}\frac15,\tan^{-1}\frac17$ cancel in pairs (telescoping), leaving: …

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4. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\theta=\text{Tan}^{-1}\left(\dfrac{1}{3}\right)+\text{Tan}^{-1}\left(\dfrac{1}{7}\right)+\text{Tan}^{-1}\left(\dfrac{1}{13}\right)+\text{Tan}^{-1}\left(\dfrac{1}{21}\right)+\text{Tan}^{-1}\left(\dfrac{1}{31}\right)$, then $\tan\theta=$ (A) $\dfrac{3}{5}$ (B) 1 (C) $\dfrac{5}{7}$ (D) $\dfrac{7}{9}$

   ›Reveal solutionSolution

   Recognizing $3,7,13,21,31$ as the sequence $n^2+n+1$ turns each $\arctan$ term into a telescoping difference $\arctan(n+1)-\arctan n$. Answer: $\tan\theta=5/7$.

   Concept and Intuition

   The telescoping arctan identity $\arctan(n+1)-\arctan n=\arctan\dfrac{1}{1+n(n+1)}=\arctan\dfrac1{n^2+n+1}$ turns a long sum of small-angle arctangents into just the first and last terms — a huge simplification once the denominators are recognized as $n^2+n+1$.

   Step-by-Step Solution

   1. Check the denominators: for $n=1,2,3,4,5$, $n^2+n+1=3,7,13,21,31$ — exactly matching the given series.
   2. Use $\tan^{-1}(n+1)-\tan^{-1}n=\tan^{-1}\dfrac{(n+1)-n}{1+n(n+1)}=\tan^{-1}\dfrac{1}{n^2+n+1}$.
   3. So $\theta=\displaystyle\sum_{n=1}^{5}\left[\tan^{-1}(n+1)-\tan^{-1}n\right]$, which telescopes: all middle terms cancel, leaving $\theta=\tan^{-1}6-\tan^{-1}1$.
   4. $\tan^{-1}1=\pi/4$, so $\theta=\tan^{-1}6-\pi/4$. …

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5. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If $y=\text{Tan}^{-1}\dfrac{x}{1+2x^2}+\text{Tan}^{-1}\dfrac{x}{1+6x^2}+\text{Tan}^{-1}\dfrac{x}{1+12x^2}$, then $\left(\dfrac{dy}{dx}\right)_{x=\frac{1}{2}}=$ (A) $1$ (B) $-1$ (C) $0$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   Recognize the telescoping arctan pattern to collapse $y$ to $\arctan(4x)-\arctan(x)$; the derivative at $x=\frac12$ is $0$.

   Concept and Intuition

   The identity $\arctan A-\arctan B=\arctan\dfrac{A-B}{1+AB}$ (mod branch issues) means a sum like $\arctan\dfrac{x}{1+n(n+1)x^2}$, recognized as $\arctan((n+1)x)-\arctan(nx)$, telescopes when summed over consecutive $n$ — a huge simplification before ever differentiating.

   Step-by-Step Solution

   1. Check the general term: $\arctan((n+1)x)-\arctan(nx)=\arctan\dfrac{(n+1)x-nx}{1+n(n+1)x^2}=\arctan\dfrac{x}{1+n(n+1)x^2}$.
   2. Match given terms: $\dfrac{x}{1+2x^2}$ has $n(n+1)=2\Rightarrow n=1$; $\dfrac{x}{1+6x^2}$ has $n(n+1)=6\Rightarrow n=2$; $\dfrac{x}{1+12x^2}$ has $n(n+1)=12\Rightarrow n=3$.
   3. So $y=[\arctan2x-\arctan x]+[\arctan3x-\arctan2x]+[\arctan4x-\arctan3x]=\arctan4x-\arctan x$ (everything else cancels). …

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6. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\lim_{n \to \infty} \sum_{r=1}^{n} \cot^{-1}\left(r^2 + \dfrac{3}{4}\right) =$ (A) $\cot^{-1} 2$ (B) $\cot^{-1} \dfrac{1}{3}$ (C) $\tan^{-1} 2$ (D) $\tan^{-1} \dfrac{1}{3}$

   ›Reveal solutionSolution

   Rewriting each term as a difference of two arctangents makes the sum telescope; taking the limit as $n\to\infty$ gives $\tan^{-1}2$.

   Concept and Intuition

   Sums of $\tan^{-1}$ or $\cot^{-1}$ terms with a quadratic argument like $r^2+c$ often telescope, because $r^2+\tfrac34 = 1+(r+\tfrac12)(r-\tfrac12)$ matches the arctangent subtraction identity $\tan^{-1}A-\tan^{-1}B=\tan^{-1}\dfrac{A-B}{1+AB}$ with $A=r+\tfrac12$, $B=r-\tfrac12$ (so $A-B=1$).

   Step-by-Step Solution

   1. $\cot^{-1}\left(r^2+\tfrac34\right)=\tan^{-1}\dfrac{1}{r^2+3/4}$.
   2. Let $A=r+\tfrac12$, $B=r-\tfrac12$: then $AB=r^2-\tfrac14$, so $1+AB=r^2+\tfrac34$, and $A-B=1$.
   3. So $\tan^{-1}\dfrac{1}{r^2+3/4}=\tan^{-1}\dfrac{A-B}{1+AB}=\tan^{-1}A-\tan^{-1}B=\tan^{-1}\!\left(r+\tfrac12\right)-\tan^{-1}\!\left(r-\tfrac12\right)$.
   4. Sum from $r=1$ to $n$ telescopes: $\sum_{r=1}^{n}\left[\tan^{-1}(r+\tfrac12)-\tan^{-1}(r-\tfrac12)\right]=\tan^{-1}(n+\tfrac12)-\tan^{-1}(\tfrac12)$. …

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7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\tan\left(2\,\mathrm{Tan}^{-1}\left(\dfrac{1}{3}\right) + \mathrm{Tan}^{-1}\left(\dfrac{1}{7}\right)\right) =$ (A) $\dfrac{1}{\sqrt{3}}$ (B) $\sqrt{3}$ (C) $1$ (D) $3/7$

   ›Reveal solutionSolution

   First reduce the double-angle inverse-tangent term to a single tangent value using the tangent double-angle formula, then combine with the second $\tan^{-1}$ term using the tangent addition formula. The result is exactly $1$.

   Concept and Intuition

   Expressions like $\tan(2\tan^{-1}x+\tan^{-1}y)$ are handled in two stages: first collapse $2\tan^{-1}x$ to a single angle whose tangent is known via the double-angle formula $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}$, then treat the whole thing as $\tan(\alpha'+\beta)$ using the standard addition formula, where $\alpha'$ is the angle with $\tan\alpha'=\tan(2\tan^{-1}x)$.

   Step-by-Step Solution

   1. Let $\alpha=\tan^{-1}(1/3)$, so $\tan\alpha=1/3$. Then $\tan2\alpha=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2/3}{1-1/9}=\dfrac{2/3}{8/9}=\dfrac{2}{3}\times\dfrac{9}{8}=\dfrac{18}{24}=\dfrac34$.
   2. Let $\beta=\tan^{-1}(1/7)$, so $\tan\beta=1/7$.
   3. We need $\tan(2\alpha+\beta)=\dfrac{\tan2\alpha+\tan\beta}{1-\tan2\alpha\tan\beta}=\dfrac{\frac34+\frac17}{1-\frac34\cdot\frac17}$. …

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8. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.$\dfrac{d}{dx}\left(\operatorname{Tan}^{-1}\left(\dfrac{\cos x}{1+\sin x}\right)\right)=$ (A) $\dfrac{1}{2}$ (B) $\dfrac{-1}{2}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   Simplify the argument to a single tangent of a half-angle expression, then differentiate. Answer: $-\dfrac12$.

   Concept and Intuition

   Expressions like $\dfrac{\cos x}{1+\sin x}$ are classic half-angle simplifications that collapse to $\tan\left(\dfrac\pi4-\dfrac x2\right)$, letting the inverse tangent cancel with the tangent directly.

   Step-by-Step Solution

   1. Multiply numerator and denominator by $(1-\sin x)$: $\dfrac{\cos x}{1+\sin x}=\dfrac{\cos x(1-\sin x)}{1-\sin^2x}=\dfrac{\cos x(1-\sin x)}{\cos^2x}=\dfrac{1-\sin x}{\cos x}$.
   2. This is a known identity: $\dfrac{1-\sin x}{\cos x}=\tan\left(\dfrac\pi4-\dfrac x2\right)$. …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $y = \mathrm{Tan}^{-1}\left(\dfrac{x}{1+2x^2}\right) + \mathrm{Tan}^{-1}\left(\dfrac{x}{1+6x^2}\right)$, then $\dfrac{dy}{dx}=$ (A) $\dfrac{4}{16x^2+1} - \dfrac{3}{9x^2+1}$ (B) $\dfrac{3}{9x^2+1} - \dfrac{1}{x^2+1}$ (C) $\dfrac{3}{9x^2+1} - \dfrac{2}{4x^2+1}$ (D) $\dfrac{1}{9x^2+1} - \dfrac{1}{x^2+1}$

   ›Reveal solutionSolution

   Recognizing each arctan term as a telescoping difference $\tan^{-1}(2x)-\tan^{-1}(x)$ and $\tan^{-1}(3x)-\tan^{-1}(2x)$ collapses $y$ to $\tan^{-1}3x-\tan^{-1}x$, whose derivative is immediate.

   Concept and Intuition

   Terms of the form $\tan^{-1}\!\left(\dfrac{a-b}{1+ab}\right)$ are exactly $\tan^{-1}a - \tan^{-1}b$ (the tangent subtraction identity, valid when $ab>-1$). Spotting this pattern turns an awkward-looking sum into a telescoping simplification, avoiding messy direct differentiation of nested rational-argument arctans.

   Step-by-Step Solution

   1. Compare $\dfrac{x}{1+2x^2}$ to the form $\dfrac{a-b}{1+ab}$: try $a=2x,\,b=x$, giving $\dfrac{2x-x}{1+2x\cdot x}=\dfrac{x}{1+2x^2}$ ✓. So

   $\tan^{-1}\!\left(\frac{x}{1+2x^2}\right) = \tan^{-1}(2x)-\tan^{-1}(x)$

   2. Compare $\dfrac{x}{1+6x^2}$ similarly: try $a=3x,\,b=2x$, giving $\dfrac{3x-2x}{1+3x\cdot2x}=\dfrac{x}{1+6x^2}$ ✓. So

   $\tan^{-1}\!\left(\frac{x}{1+6x^2}\right) = \tan^{-1}(3x)-\tan^{-1}(2x)$

   3. Add the two:

   $y = \big[\tan^{-1}2x-\tan^{-1}x\big] + \big[\tan^{-1}3x-\tan^{-1}2x\big] = \tan^{-1}3x - \tan^{-1}x$

   (the $\tan^{-1}2x$ terms cancel — a telescoping sum).

   4. Differentiate: …

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10. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.If $y = \mathrm{Tan}^{-1}\left\{\dfrac{ax-b}{bx+a}\right\}$, then $y' =$ _______ (A) $\dfrac{1}{1+x^2} + \dfrac{a^2}{a^2+b^2}$ (B) $\dfrac{1}{1+x^2}$ (C) $\dfrac{1}{1+\left(\dfrac{ax-b}{bx+a}\right)^2}$ (D) $\dfrac{bx+a}{1+(ax-b)^2}$

    ›Reveal solutionSolution

    This tests differentiating an arctangent of a Möbius-type expression, which (as often happens) collapses to the simple form $\tfrac{1}{1+x^2}$. Answer: $\dfrac{1}{1+x^2}$.

    Concept and Intuition

    Expressions of the form $\mathrm{Tan}^{-1}\!\left(\dfrac{ax-b}{bx+a}\right)$ often equal $\mathrm{Tan}^{-1}(x) - \mathrm{Tan}^{-1}(b/a)$ up to a constant (since $\dfrac{ax-b}{bx+a} = \dfrac{\frac{a}{b}x - 1}{x+\frac ab}$ resembles $\tan(\alpha-\beta)$ with a constant angle $\beta$), so its derivative should reduce to just $\dfrac{1}{1+x^2}$ — the constant angle contributes zero derivative.

    Step-by-Step Solution

    1. Let $u = \dfrac{ax-b}{bx+a}$. By the quotient rule, $u' = \dfrac{a(bx+a) - (ax-b)b}{(bx+a)^2} = \dfrac{abx+a^2-abx+b^2}{(bx+a)^2} = \dfrac{a^2+b^2}{(bx+a)^2}$.
    2. Compute $1+u^2 = 1+\dfrac{(ax-b)^2}{(bx+a)^2} = \dfrac{(bx+a)^2+(ax-b)^2}{(bx+a)^2}$. …

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11. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If $\theta = 2\tan^{-1}\dfrac{1}{8} + 2\tan^{-1}\dfrac{1}{5} + \tan^{-1}\dfrac{1}{7}$ and $\tan\dfrac{\theta}{2} = \sqrt{m} + \sqrt{n}$ where $m$ and $n$ are positive integers such that $m < n$ then $(m^n + n^m)^{m+n} =$ (A) $18$ (B) $27$ (C) $25$ (D) $36$

    ›Reveal solutionSolution

    Simplifying the sum of inverse tangents gives $\theta=\pi/4$, so $\tan(\theta/2)=\tan(\pi/8)=\sqrt2-1$; with $m=1,n=2$ this gives $(m^n+n^m)^{m+n}=27$.

    Concept and Intuition

    Repeated use of the tangent addition formula $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\!\left(\dfrac{a+b}{1-ab}\right)$ collapses the sum of arctangents into a single, simple angle.

    Step-by-Step Solution

    1. $\tan^{-1}\dfrac18+\tan^{-1}\dfrac15 = \tan^{-1}\!\left(\dfrac{1/8+1/5}{1-1/40}\right)=\tan^{-1}\!\left(\dfrac{13/40}{39/40}\right)=\tan^{-1}\dfrac13$.
    2. So $2\tan^{-1}\dfrac18+2\tan^{-1}\dfrac15 = 2\tan^{-1}\dfrac13$.
    3. Double-angle: $\tan\left(2\tan^{-1}\tfrac13\right)=\dfrac{2\cdot\frac13}{1-\frac19}=\dfrac{2/3}{8/9}=\dfrac34$, so $2\tan^{-1}\tfrac13=\tan^{-1}\tfrac34$.
    4. $\theta=\tan^{-1}\dfrac34+\tan^{-1}\dfrac17 = \tan^{-1}\!\left(\dfrac{3/4+1/7}{1-3/28}\right)=\tan^{-1}\!\left(\dfrac{25/28}{25/28}\right)=\tan^{-1}(1)=\dfrac{\pi}{4}$. …

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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.$4\tan^{-1}\dfrac{1}{5} - \tan^{-1}\dfrac{1}{70} + \tan^{-1}\dfrac{1}{99} =$ (A) $\dfrac{\pi}{12}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{3}$

    ›Reveal solutionSolution

    This is a Machin-like arctangent identity that evaluates exactly to $\pi/4$.

    Concept and Intuition

    Sums/differences of $\tan^{-1}$ terms with small reciprocal arguments often combine (via repeated use of $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\frac{p-q}{1+pq}$ and the double/quadruple-angle formula for tangent) into a single nice angle like $\pi/4$. These are the classical 'Machin-type' formulas historically used to compute $\pi$.

    Step-by-Step Solution

    1. First combine $4\tan^{-1}\frac15$ using the double-angle formula for $\tan$ twice: with $\tan\alpha=\frac15$, $\tan2\alpha=\dfrac{2\cdot\frac15}{1-\frac1{25}}=\dfrac{2/5}{24/25}=\dfrac{5}{12}$, and $\tan4\alpha=\dfrac{2\cdot\frac5{12}}{1-\frac{25}{144}}=\dfrac{5/6}{119/144}=\dfrac{120}{119}$.
    2. So $4\tan^{-1}\frac15 = \tan^{-1}\dfrac{120}{119}$ (in the correct quadrant, since $\frac{120}{119}$ is only slightly bigger than 1, the angle is just over $\pi/4$).
    3. Now combine $\tan^{-1}\dfrac{120}{119}-\tan^{-1}\dfrac1{70}$ using $\tan^{-1}p-\tan^{-1}q=\tan^{-1}\frac{p-q}{1+pq}$: numerator $\frac{120}{119}-\frac1{70} = \frac{120\cdot70-119}{119\cdot70}=\frac{8400-119}{8330}=\frac{8281}{8330}$; denominator $1+\frac{120}{119\cdot70}=1+\frac{120}{8330}=\frac{8450}{8330}$. Ratio $=\frac{8281}{8450}$. …

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