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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

With usual notations in Δ ABC , if C=90^ ^ ^ , then ^ -1...

With usual notations in ΔABC\Delta \mathrm{ABC}, if C=90∘C=90^{{}^{{}^{\circ}}}, then tan⁡−1\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}(bc+a)=\left(\frac{b}{c+a}\right)=With usual notations in Δ ABC , if C=90^ ^ ^ , then ^ -1 ((a)/(b+c))+ ^ -1 ((b)/(c+a))=

Options

  1. Aπ4\frac{\pi}{4}
  2. Bπ6\frac{\pi}{6}
  3. Cπ\pi
  4. Dπ3\frac{\pi}{3}

Correct answer

A. π4\frac{\pi}{4}

Step-by-step solution

∴tan⁡−1(ab+c)+tan⁡−1(bc+a)=tan⁡−1[ab+c+bc+a1−(ab+c)(bc+a)]=tan⁡−1[ac+a2+bc+b2ac+bc+ab+c2−ab]=tan⁡−1(ac+bc+c2ac+bc+c2)…[∵a2+b2=c2\therefore \quad \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]=\tan ^{-1}\left[\frac{a c+a^{2}+b c+b^{2}}{a c+b c+a b+c^{2}-a b}\right]=\tan ^{-1}\left(\frac{a c+b c+c^{2}}{a c+b c+c^{2}}\right) \quad \ldots\left[\because a^{2}+b^{2}=c^{2}\right., refer diagram ]=tan⁡−1(1)=π4]=\tan ^{-1}(1)=\frac{\pi}{4} With usual notations in Δ ABC , if C=90^ ^ ^ , then ^ -1 ((a)/(b+c))+ ^ -1 ((b)/(c+a))=
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