JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(0 < a , b < 1,\) and \(\tan ^{-1} a +\tan ^{-1} b =\frac{\pi}{4},\) then the value of \((a+b)-\left(\frac{a^{2}+b^{2}}{2}\right)+\left(\frac{a^{3}+b^{3}}{3}\right)-\left(\frac{a^{4}+b^{4}}{4}\right)+\ldots\) is ..... .
- A \(\log _{ e } 2\)
- B \(e ^{2}-1\)
- C \(e\)
- D \(\log _{ e }\left(\frac{ e }{2}\right)\)
Answer & Solution
Correct Answer
(A) \(\log _{ e } 2\)
Step-by-step Solution
Detailed explanation
\(\tan ^{-1} a +\tan ^{-1} b =\frac{\pi}{4} \quad 0< a , b <1\) \(\Rightarrow \frac{a+b}{1-a b}=1\) \(a+b=1-a b\) \((a+1)(b+1)=2\) Now \(\left[ a -\frac{ a ^{2}}{2}+\frac{ a ^{3}}{3}+\ldots\right]+\left[ b -\frac{ b ^{2}}{2}+\frac{ b ^{3}}{3}+\ldots\right]\)…
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