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

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.What is the value of $\mathrm{Sin}^{-1}\dfrac{12}{13} + \mathrm{Cos}^{-1}\dfrac{4}{5} + \mathrm{Tan}^{-1}\dfrac{63}{16} =$ (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{3\pi}{4}$

   ›Reveal solutionSolution

   The three inverse-trig terms telescope to exactly $\pi$ because the first two sum to $180^\circ$ minus the third.

   Concept and Intuition

   $\sin^{-1}\frac{12}{13}$ is the acute angle with sine $\frac{12}{13}$, cosine $\frac{5}{13}$, tangent $\frac{12}{5}$. $\cos^{-1}\frac{4}{5}$ is the acute angle with cosine $\frac{4}{5}$, sine $\frac{3}{5}$, tangent $\frac{3}{4}$. Adding two acute angles whose tangent-sum formula gives a negative tangent tells us their sum exceeds $90^\circ$ — a classic trick for handling sums of inverse trig terms without a calculator.

   Step-by-Step Solution

   1. Let $\alpha=\sin^{-1}\frac{12}{13}$ (so $\tan\alpha=\frac{12}{5}$) and $\beta=\cos^{-1}\frac{4}{5}$ (so $\tan\beta=\frac{3}{4}$).
   2. $\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}=\dfrac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\cdot\frac{3}{4}}=\dfrac{\frac{63}{20}}{-\frac{16}{20}}=-\dfrac{63}{16}$.
   3. Since $\alpha\approx 67.4^\circ,\beta\approx 36.9^\circ$, their sum $\alpha+\beta\approx 104.3^\circ\in(90^\circ,180^\circ)$, where tangent is negative — consistent. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $=\frac{8281}{8450}$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\sin\left[2\tan^{-1}\left(\dfrac{1}{2}\right) + \sin^{-1}\left(\dfrac{3}{5}\right)\right] =$ (A) 0 (B) 1 (C) $\dfrac{1}{2}$ (D) $\dfrac{\sqrt3}{2}$

   ›Reveal solutionSolution

   Convert both inverse-trig terms into a single angle's sine/cosine using right-triangle ratios, then apply the sine addition formula; the sum evaluates neatly to $1$.

   Concept and Intuition

   When adding two inverse trig angles, the cleanest approach is to name each one, extract its sine and cosine from the implied right triangle, then use the standard addition formula $\sin(A+B)=\sin A\cos B+\cos A\sin B$ — never try to add the angles numerically.

   Step-by-Step Solution

   1. Let $\varphi = \tan^{-1}\left(\dfrac12\right)$. In a right triangle, opposite $=1$, adjacent $=2$, hypotenuse $=\sqrt5$.
   2. Use the double-angle identity $\tan2\varphi = \dfrac{2\tan\varphi}{1-\tan^2\varphi} = \dfrac{2(1/2)}{1-(1/4)} = \dfrac{1}{3/4}=\dfrac43$.
   3. Since $\varphi\in(0,\pi/4)$ (as $\tan\varphi=1/2<1$), $2\varphi\in(0,\pi/2)$, so this is a genuine first-quadrant angle: with opposite $4$, adjacent $3$, hypotenuse $5$, giving $\sin2\varphi=4/5,\ \cos2\varphi=3/5$.
   4. Let $\psi=\sin^{-1}(3/5)$, so $\sin\psi=3/5$ and (principal branch, first quadrant) $\cos\psi=4/5$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\tan^{-1}\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{15}+1} + \tan^{-1}\dfrac{1}{\sqrt{5}} =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   This tests simplifying a nested surd inside an inverse trig function, then adding two inverse-tangent values. The answer is (A).

   Concept and Intuition

   A nested surd of the form $\sqrt{a+b-2\sqrt{ab}}$ always simplifies to $|\sqrt a-\sqrt b|$. Recognizing $8-2\sqrt{15}$ as this form with $a=5,b=3$ turns an ugly expression into a clean one, after which the sum of the two arctangents can be identified as a standard angle.

   Step-by-Step Solution

   1. Write $8-2\sqrt{15}=5+3-2\sqrt{5\cdot3}=(\sqrt5-\sqrt3)^2$, so $\sqrt{8-2\sqrt{15}}=\sqrt5-\sqrt3$ (positive since $\sqrt5>\sqrt3$).
   2. The first term becomes $\tan^{-1}\dfrac{\sqrt5-\sqrt3}{\sqrt{15}+1}$.
   3. Numerically: $\sqrt5\approx2.236$, $\sqrt3\approx1.732$, $\sqrt{15}\approx3.873$. So the argument $\approx\dfrac{0.504}{4.873}\approx0.1034$, giving the first term $\approx5.91°$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\sin\left(2\tan^{-1}\left(\dfrac{1}{3}\right)\right) + \cos\left(\tan^{-1}2\sqrt{2}\right) =$ (A) $\dfrac{16}{15}$ (B) $\dfrac{14}{15}$ (C) $\dfrac{11}{15}$ (D) $\dfrac{8}{15}$

   ›Reveal solutionSolution

   This tests reading sine/cosine of an inverse-tangent angle off a right triangle, then applying the double-angle sine formula. The sum evaluates to $14/15$.

   Concept and Intuition

   Given $\tan^{-1}(p/q)$, build the right triangle with opposite $p$, adjacent $q$, hypotenuse $\sqrt{p^2+q^2}$; then any trig function of that angle is a direct ratio of the triangle's sides. This avoids working with inverse functions directly.

   Step-by-Step Solution

   1. Let $\alpha=\tan^{-1}(1/3)$: right triangle with opposite $1$, adjacent $3$, hypotenuse $\sqrt{1+9}=\sqrt{10}$. So $\sin\alpha=\dfrac{1}{\sqrt{10}}$, $\cos\alpha=\dfrac{3}{\sqrt{10}}$.
   2. $\sin2\alpha = 2\sin\alpha\cos\alpha = 2\cdot\dfrac{1}{\sqrt{10}}\cdot\dfrac{3}{\sqrt{10}} = \dfrac{6}{10}=\dfrac{3}{5}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\sin\left(\text{Tan}^{-1}\dfrac{12}{17} + \text{Tan}^{-1}\dfrac{5}{29}\right) =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $\dfrac{\sqrt{3}}{2}$ (D) $\dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   This tests the sine-addition formula built from two inverse-tangent right triangles; the arithmetic collapses very neatly since $866=2\times433$.

   Concept and Intuition

   $\tan^{-1}\frac{12}{17}$ is the angle of a right triangle with opposite $12$, adjacent $17$, hypotenuse $\sqrt{12^2+17^2}=\sqrt{433}$. Similarly $\tan^{-1}\frac{5}{29}$ gives a triangle with hypotenuse $\sqrt{5^2+29^2}=\sqrt{866}$. Once we have $\sin$ and $\cos$ of both angles, the sum formula does the rest — no need to ever find the angles themselves.

   Step-by-Step Solution

   1. Let $A=\tan^{-1}\frac{12}{17}$: $\sin A=\dfrac{12}{\sqrt{433}}$, $\cos A=\dfrac{17}{\sqrt{433}}$ (since $12^2+17^2=144+289=433$).
   2. Let $B=\tan^{-1}\frac{5}{29}$: $\sin B=\dfrac{5}{\sqrt{866}}$, $\cos B=\dfrac{29}{\sqrt{866}}$ (since $5^2+29^2=25+841=866$).
   3. $\sin(A+B)=\sin A\cos B+\cos A\sin B=\dfrac{12(29)+17(5)}{\sqrt{433}\sqrt{866}}=\dfrac{348+85}{\sqrt{433\cdot866}}=\dfrac{433}{\sqrt{433\cdot866}}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\mathrm{Tan}^{-1} 2 + \mathrm{Tan}^{-1} 3 =$ (A) $-\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{3\pi}{4}$ (D) $\dfrac{5\pi}{4}$

   ›Reveal solutionSolution

   Adding two inverse-tangent principal values whose tangent-sum formula gives $-1$, the actual sum is $3\pi/4$ (not $-\pi/4$), because both individual angles are obtuse-leaning acute angles summing past $\pi/2$.

   Concept and Intuition

   The tangent addition formula only gives $\tan(A+B)$, not $A+B$ directly — since tangent is periodic with period $\pi$, we must use the actual sizes of $A=\tan^{-1}2$ and $B=\tan^{-1}3$ (each in $(0,\pi/2)$, and in fact each $>\pi/4$ since $\tan>1$) to determine which branch the sum falls into.

   Step-by-Step Solution

   1. Let $A=\tan^{-1}2$, $B=\tan^{-1}3$; both lie in $(\pi/4,\pi/2)$ since $\tan A=2>1,\tan B=3>1$.
   2. $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{2+3}{1-6}=\dfrac{5}{-5}=-1$.
   3. Since $A,B\in(\pi/4,\pi/2)$, their sum $A+B\in(\pi/2,\pi)$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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: …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $\theta=\tan^{-1}6-\pi/4$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.$\mathrm{Tanh}^{-1}\left(\dfrac{1}{3}\right) + \mathrm{Coth}^{-1}(3) =$ (A) $\mathrm{Sech}^{-1}\left(\dfrac{1}{3}\right)$ (B) $\mathrm{Cosech}^{-1}\left(\dfrac{1}{3}\right)$ (C) $\mathrm{Cosh}^{-1}\left(\dfrac{4}{3}\right)$ (D) $\mathrm{Sinh}^{-1}\left(\dfrac{3}{4}\right)$

    ›Reveal solutionSolution

    This tests the identity linking $\mathrm{Coth}^{-1}$ to $\mathrm{Tanh}^{-1}$ and the logarithmic form of inverse hyperbolic functions; the sum collapses to $\log 2 = \mathrm{Sinh}^{-1}(3/4)$.

    Concept and Intuition

    Inverse hyperbolic functions all reduce to logarithms. For $|x|>1$, $\mathrm{Coth}^{-1}(x) = \mathrm{Tanh}^{-1}(1/x)$ because $\coth\theta = x \iff \tanh\theta = 1/x$. This lets us rewrite both terms of the sum using the SAME inverse function, so they simply add.

    Step-by-Step Solution

    1. Since $3>1$, use $\mathrm{Coth}^{-1}(3) = \mathrm{Tanh}^{-1}(1/3)$.
    2. The sum becomes $\mathrm{Tanh}^{-1}(1/3) + \mathrm{Tanh}^{-1}(1/3) = 2\,\mathrm{Tanh}^{-1}(1/3)$.
    3. Use $\mathrm{Tanh}^{-1}(y) = \dfrac12 \log\!\left(\dfrac{1+y}{1-y}\right)$ with $y = 1/3$: $\mathrm{Tanh}^{-1}(1/3) = \dfrac12\log\!\left(\dfrac{4/3}{2/3}\right) = \dfrac12 \log 2$.
    4. So the sum $= 2\times \dfrac12\ln2 = \log 2$.
    5. Test $\mathrm{Sinh}^{-1}(3/4) = \log\!\left(y+\sqrt{y^2+1}\right)$ with $y=3/4$: $\log\!\left(\dfrac34+\sqrt{\dfrac{9}{16}+1}\right) = \log\!\left(\dfrac34+\dfrac54\right) = \log 2$. Exact match. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If $\dfrac{1}{2}\sin^{-1}\left(\dfrac{3\sin 2\theta}{5+4\cos 2\theta}\right) = \tan^{-1}x$ then $x =$ (A) $\tan\dfrac{\theta}{3}$ (B) $\dfrac{1}{3}\tan\theta$ (C) $\tan 3\theta$ (D) $\dfrac{1}{3}\tan 3\theta$

    ›Reveal solutionSolution

    Rewrite the inverse-sine argument as $\sin(2\phi)$ for a suitable $\tan\phi$ built from $\tan\theta$, so the half of the inverse sine collapses to $\tan^{-1}$ of that quantity directly. Answer: $x=\dfrac13\tan\theta$.

    Concept and Intuition

    Expressions of the form $\dfrac{k\sin2\theta}{A+B\cos2\theta}$ can often be massaged, after dividing through by $(1+\tan^2\theta)$ i.e. converting to $t=\tan\theta$, into the recognizable double-angle sine pattern $\dfrac{2u}{1+u^2}=\sin(2\tan^{-1}u)$ for some rescaled variable $u$. Spotting this pattern converts an awkward inverse-trig expression into a clean angle.

    Step-by-Step Solution

    1. Let $t=\tan\theta$. Recall $\sin2\theta=\dfrac{2t}{1+t^2}$, $\cos2\theta=\dfrac{1-t^2}{1+t^2}$.
    2. Numerator: $3\sin2\theta=\dfrac{6t}{1+t^2}$.
    3. Denominator: $5+4\cos2\theta=5+\dfrac{4(1-t^2)}{1+t^2}=\dfrac{5(1+t^2)+4(1-t^2)}{1+t^2}=\dfrac{9+t^2}{1+t^2}$.
    4. So the ratio is $\dfrac{6t/(1+t^2)}{(9+t^2)/(1+t^2)}=\dfrac{6t}{9+t^2}$.
    5. Let $u=t/3$. Then $\dfrac{2u}{1+u^2}=\dfrac{2t/3}{1+t^2/9}=\dfrac{2t/3}{(9+t^2)/9}=\dfrac{6t}{9+t^2}$ — exactly matches!
    6. So $\dfrac{6t}{9+t^2}=\sin(2\phi)$ where $\tan\phi=u=t/3$, i.e. $\phi=\tan^{-1}\!\left(\dfrac{\tan\theta}{3}\right)$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 $\tan(2\alpha+\beta)=\dfrac{\tan2\alpha+\tan\beta}{1-\tan2\alpha\tan\beta}=\dfrac{\frac34+\frac17}{1-\frac34\cdot\frac17}$. …

    🔒 Start your 14-day free trial to see the full solution →