Q.8 If tan⁻¹(2x)+tan⁻¹(3x)=π/4 then find the value of x , where 0lt 3xlt 1

Answer: x=1/6 Since 0lt 3xlt1 , 2x·3x=6x²lt1 , so tan⁻¹(2x)+tan⁻¹(3x)=tan⁻¹(2x+3x/1-6x²)=π/4 . therefore 5x/1-6x²=tanπ/4=1⇒ 6x²+5x-1=0⇒(6x-1)(x+1)=0 . x=1/6 or x=-1 . As 0lt3xlt1 , x=-1 is rejected. Hence x=1/6 .

Chapter: Trigonometric Functions· Mathematics & Statistics Part 1

If \(\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}\) then find the value of \(x\), where \(0\lt 3x\lt 1\)

Q.8 If \(\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}\) then find the value of \(x\), where \(0\lt 3x\lt 1\)

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

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Answer: \(x=\dfrac16\)

Since \(0\lt 3x\lt1\), \(2x\cdot3x=6x^2\lt1\), so \(\tan^{-1}(2x)+\tan^{-1}(3x)=\tan^{-1}\left(\dfrac{2x+3x}{1-6x^2}\right)=\dfrac{\pi}{4}\).

\(\therefore \dfrac{5x}{1-6x^2}=\tan\dfrac{\pi}{4}=1\Rightarrow 6x^2+5x-1=0\Rightarrow(6x-1)(x+1)=0\).

\(x=\dfrac16\) or \(x=-1\). As \(0\lt3x\lt1\), \(x=-1\) is rejected. Hence \(x=\dfrac16\).

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

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If \(\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}\) then find the value of \(x\), where \(0\lt 3x\lt 1\)

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