Question
Download Solution PDF\(\tan ^{-1}\left(\frac{a}{b}\right)-\tan ^{-1}\left(\frac{a-b}{a+b}\right)\) किसके बराबर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
सूत्र: tan-1x - tan-1y = tan-1 \(x-y\over 1+xy\)
स्पष्टीकरण:
\(\tan ^{-1}\left(\frac{a}{b}\right)-\tan ^{-1}\left(\frac{a-b}{a+b}\right)\)
= \(\tan ^{-1}\left({\frac{a}{b}-{a-b\over a+b}\over1+\frac ab.{a-b\over a+b}}\right)\)
= \(\tan ^{-1}\left({\frac{a^2+ab-ab+b^2}{b(a+b)}\over{ab+b^2+a^2-ab\over b(a+b)}}\right)\)
= tan -1 \(a^2+b^2\over a^2+b^2\) = tan -1 1 = \(\frac{\pi}{4}\)
विकल्प (2) सत्य है।
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