Considering only the principal values of inverse trigonometric functions, the number of positive rea

Q: Why can't I just use the $\tan^{-1}(A) + \tan^{-1}(B)$ formula directly?

You absolutely can! However, that formula is defined as $\tan^{-1}(\frac{A+B}{1-AB})$ only when $AB 1$, you get an extra $\pi$ term. Rearranging the equation first avoids this ambiguity entirely, making your path to the solution much safer and more elegant.

Q: Why do we reject the negative root?

The problem explicitly states that we are looking for positive real values of $x$. When we solve the quadratic equation $2x^2 + 3x - 1 = 0$, we get two roots. One is negative, and one is positive. Since the negative root violates the initial constraint $x > 0$, it is mathematically irrelevant to the specific question asked.

Q: Is there a graphical way to solve this?

Yes! If you plot $f(x) = \tan^{-1}(x) + \tan^{-1}(2x)$, you will see it is a strictly increasing function starting from $(0,0)$. Since it is strictly increasing, it can intersect the horizontal line $y = \frac{\pi}{4}$ at most once. This confirms our algebraic result that there is exactly one solution.

Question

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $	an^{-1}(x) + 	an^{-1}(2x) = \frac{\pi}{4}$ is :

Options:

(A) More than 2

(B) 1

(C) 2

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup We are tasked with solving the equation $ an^{-1}(x) + an^{-1}(2x) = \frac{\pi}{4}$. At first glance, this appears to be a standard inverse trigonometric identity problem. However, as any seasoned JEE aspirant knows, the devil is in the details—specifically, the constraint $x > 0$. Imagine you are standing on the graph of $f(x) = an^{-1}(x) + an^{-1}(2x)$. As $x$ increases from zero, both $ an^{-1}(x)$ and $ an^{-1}(2x)$ grow steadily, meaning our function $f(x)$ is strictly increasing. We are looking for the unique point where this curve hits the horizontal line $y = \frac{\pi}{4}$. ### The Strategic Shift You might be tempted to use the standard addition formula $ an^{-1}(A) + an^{-1}(B) = an^{-1}(\frac{A+B}{1-AB})$. However, this formula is a trap because it behaves differently depending on whether $AB < 1$ or $AB > 1$. To avoid this pitfall, we rearrange the equation to isolate the inverse functions: $$ an^{-1}(2x) = \frac{\pi}{4} - an^{-1}(x)$$ By doing this, we can apply the tangent function to both sides without worrying about the $AB < 1$ condition. Applying the tangent function yields: $$ an( an^{-1}(2x)) = an\left(\frac{\pi}{4} - an^{-1}(x) ight)$$ ### Solving the Algebraic Equation On the left, the tangent and inverse tangent cancel out, leaving us with $2x$. On the right, we use the identity $ an(A - B) = \frac{ an A - an B}{1 + an A an B}$. Setting $A = \frac{\pi}{4}$ and $B = an^{-1}(x)$, the right side becomes: $$\frac{ an(\frac{\pi}{4}) - an( an^{-1}(x))}{1 + an(\frac{\pi}{4}) an( an^{-1}(x))} = \frac{1 - x}{1 + x}$$ This leads to the simple algebraic equation: $$2x = \frac{1 - x}{1 + x}$$ Cross-multiplying gives $2x(1 + x) = 1 - x$, which expands to $2x^2 + 2x = 1 - x$. Rearranging terms, we arrive at the quadratic equation: $$2x^2 + 3x - 1 = 0$$ ### Final Calculation Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find: $$x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-1)}}{4} = \frac{-3 \pm \sqrt{17}}{4}$$ We have two potential roots: $x_1 = \frac{-3 - \sqrt{17}}{4}$ and $x_2 = \frac{-3 + \sqrt{17}}{4}$. Recalling our constraint $x > 0$, we observe that the first root is negative. The second root is positive because $\sqrt{17} > 4$. Thus, the only valid solution is: **$$x = \frac{\sqrt{17} - 3}{4}$$**

Animated Solution for Mathematics - Inverse Trigonometric Functions: Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ is :

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Visualized Solution

$tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$

$tan^{- 1} (2 x) = \frac{π}{4} - tan^{- 1} (x)$

$tan (tan^{- 1} (2 x)) = tan (\frac{π}{4} - tan^{- 1} (x))$

$tan (A - B) = \frac{t a n A - t a n B}{1 + t a n A t a n B}$

$2 x = \frac{1 - x}{1 + x}$

$2 x (1 + x) = 1 - x$

$2 x^{2} + 3 x - 1 = 0$

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

$x = \frac{- 3 \pm 17}{4}$

$x > 0$

$x = \frac{17 - 3}{4}$

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Strategic Shift

Solving the Algebraic Equation

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1(x)+tan−1(2x)=4π​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1(2x)=4π​−tan−1(x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(tan−1(2x))=tan(4π​−tan−1(x))

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(A−B)=1+tanAtanBtanA−tanB​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x=1+x1−x​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x(1+x)=1−x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2x2+3x−1=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=2a−b±b2−4ac​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=4−3±17​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x>0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=417​−3​