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JEE Main 2024
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Hard

Question

If cot -1 (α\alpha) = cot -1 2 + cot -1 8 + cot -1 18 + cot -1 32 + ...... upto 100 terms, then α\alpha is :

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Solution

cot1(α)=cot12+cot18+cot118+cot132+....100{\cot ^{ - 1}}(\alpha ) = co{t^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....100 terms =tan112+tan118+tan1118+tan1132+....100 = {\tan ^{ - 1}}{1 \over 2} + {\tan ^{ - 1}}{1 \over 8} + {\tan ^{ - 1}}{1 \over {18}} + {\tan ^{ - 1}}{1 \over {32}} + ....100 term =k=1100tan112k2= \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{1 \over {2{k^2}}}} =k=1100tan124k2=k=1ntan1(2k+1)(2k1)1+(2k1)(2k+1)= \sum\limits_{k = 1}^{100} {{{\tan }^{ - 1}}{2 \over {4{k^2}}} = \sum\limits_{k = 1}^n {{{\tan }^{ - 1}}{{(2k + 1) - (2k - 1)} \over {1 + (2k - 1)(2k + 1)}}} } =k=1100(tan1(2k+1)tan1(2k1))= \sum\limits_{k = 1}^{100} {\left( {{{\tan }^{ - 1}}(2k + 1) - {{\tan }^{ - 1}}(2k - 1)} \right)} =tan1201tan11 = {\tan ^{ - 1}}201 - {\tan ^{ - 1}}1 =tan1200202 = {\tan ^{ - 1}}{{200} \over {202}} =cot1(1.01) = {\cot ^{ - 1}}(1.01) Hence α=1.01\alpha = 1.01

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