Prove that $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}\left[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right]$.

Prove that tan⁻¹x + tan⁻¹y + tan⁻¹z = tan⁻¹[(x+y+z-xyz)/(1-xy-yz-zx)].

Exercise 4.5 · Q5

\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}\left[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right]

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Concept understanding — Composite and Sum/Difference Identities of Inverse Trigonometric Functions

These are the working-formula properties (Properties VI–X) for combining or composing inverse trig functions, together with the reference-triangle technique for a raw composite like $\tan(\sin^{-1}x)$ .

Composing a trig function with an unrelated inverse trig function (reference triangle). To evaluate $f(g^{-1}(x))$ where $f\ne g$ (e.g. $\cot(\sin^{-1}x)$ ), let $\theta=g^{-1}(x)$ , build a right triangle encoding $\theta$ from the definition of $g^{-1}$ (e.g. $\sin\theta=x$ gives opposite $=x$ , hypotenuse $=1$ , so adjacent $=\sqrt{1-x^2}$ by Pythagoras — taking the adjacent side non-negative since $\theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]$ keeps cosine $\ge0$ there), then read $f(\theta)$ straight off the triangle. This proves, e.g., $\tan(\sin^{-1}x)=\dfrac{x}{\sqrt{1-x^2}}$ , $-1<x<1$ .

Property VI (addition/subtraction formulas).

\sin^{-1}x+\sin^{-1}y=\sin^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right),\quad\text{if } x^2+y^2\le1 \text{ or } xy<0

\cos^{-1}x+\cos^{-1}y=\cos^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right),\quad\text{if } x+y\ge0

\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right),\quad\text{if } xy<1 \qquad \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\dfrac{x-y}{1+xy}\right),\quad\text{if } xy>-1

with analogous subtraction forms for sine and cosine. Extending the tangent addition formula to three terms gives $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\tan^{-1}\left[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right]$ ; setting the left side equal to $\pi$ and taking the tangent of both sides (which is $0$ ) proves the classical identity $x+y+z=xyz$ whenever $\tan^{-1}x+\tan^{-1}y+\tan^{-1}z=\pi$ .

Property VII (double-angle-style formulas, from setting $y=x$ in Property VI).

2\tan^{-1}x=\tan^{-1}\left(\dfrac{2x}{1-x^2}\right),\ |x|<1 \qquad 2\tan^{-1}x=\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right),\ x\ge0 \qquad 2\tan^{-1}x=\sin^{-1}\left(\dfrac{2x}{1+x^2}\right),\ |x|\le1

Property VIII. $\sin^{-1}(2x\sqrt{1-x^2})=2\sin^{-1}x$ for $|x|\le\tfrac1{\sqrt2}$ , and $\sin^{-1}(2x\sqrt{1-x^2})=2\cos^{-1}x$ for $\tfrac1{\sqrt2}\le x\le1$ (the SAME left side splits into two different right sides depending on which half of $[-1,1]$ , i.e. which principal-range piece, $x$ falls in). …

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Let $A=\tan^{-1}x,\ B=\tan^{-1}y,\ C=\tan^{-1}z$ ; combine $A+B$ using the tangent addition formula, then combine that with $C$ the same way, and simplify the resulting fraction.

Step 1. Set $A=\tan^{-1}x,\ B=\tan^{-1}y,\ C=\tan^{-1}z$ , so $\tan A=x,\ \tan B=y,\ \tan C=z$ .

Step 2. Combine $A+B$ . $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{x+y}{1-xy}$ .

Step 3. Combine $(A+B)+C$ . $\tan(A+B+C)=\dfrac{\tan(A+B)+\tan C}{1-\tan(A+B)\tan C}=\dfrac{\frac{x+y}{1-xy}+z}{1-\frac{x+y}{1-xy}\cdot z}$ .

Step 4. Clear the inner fraction by multiplying numerator and denominator by $(1-xy)$ . Numerator: $(x+y)+z(1-xy)=x+y+z-xyz$ . Denominator: $(1-xy)-z(x+y)=1-xy-zx-zy$ .

Step 5. Assemble the result. $\tan(A+B+C)=\dfrac{x+y+z-xyz}{1-xy-yz-zx}$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set ANNUAL1 markMCQ
$\tan^{-1}\left(\dfrac{1}{4}\right) + \tan^{-1}\left(\dfrac{2}{9}\right)$
(a) $\tan^{-1}\left(\dfrac{1}{2}\right)$
(b) $\dfrac{1}{2}\cos^{-1}\left(\dfrac{3}{5}\right)$
(c) $\dfrac{1}{2}\sin^{-1}\left(\dfrac{3}{5}\right)$
(d) $\dfrac{1}{2}\tan^{-1}\left(\dfrac{3}{5}\right)$
›Reveal solutionSolution
Combining the two inverse tangents gives $\tan^{-1}(1/2)$ , which by the half-angle tangent formula equals $\dfrac12\cos^{-1}\!\left(\dfrac35\right)$ .

Use the addition formula $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\dfrac{x+y}{1-xy}\right)$ , valid here since $x=\dfrac14,\ y=\dfrac29$ give $xy=\dfrac{2}{36}=\dfrac{1}{18}<1$ .

Compute $x+y=\dfrac14+\dfrac29=\dfrac{9}{36}+\dfrac{8}{36}=\dfrac{17}{36}$ .

Compute $1-xy=1-\dfrac{1}{18}=\dfrac{17}{18}$ .

So $\dfrac{x+y}{1-xy}=\dfrac{17/36}{17/18}=\dfrac{17}{36}\times\dfrac{18}{17}=\dfrac{18}{36}=\dfrac12$ .

Hence $\tan^{-1}\!\left(\dfrac14\right)+\tan^{-1}\!\left(\dfrac29\right)=\tan^{-1}\!\left(\dfrac12\right)$ .

Now express this using $\cos^{-1}(3/5)$ : let $\theta=\cos^{-1}\!\left(\dfrac35\right)$ , so $\cos\theta=\dfrac35$ and, since $\theta$ is acute, $\sin\theta=\sqrt{1-\left(\frac35\right)^2}=\dfrac45$ .

Use the half-angle formula $\tan\dfrac{\theta}{2}=\dfrac{\sin\theta}{1+\cos\theta}=\dfrac{4/5}{1+3/5}=\dfrac{4/5}{8/5}=\dfrac12$ . …
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Q.Prove that tan⁡−1x+tan⁡−1y+tan⁡−1z=tan⁡−1[x+y+z−xyz1−xy−yz−zx]\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}\left[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right]tan−1x+tan−1y+tan−1z=tan−1[1−xy−yz−zxx+y+z−xyz​].

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Q.Prove that $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \tan^{-1}\left[\dfrac{x+y+z-xyz}{1-xy-yz-zx}\right]$ .

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