Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Miscellaneous

Misc 1
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Misc 2

Misc 3

Misc 4 Important

Misc 5 Important

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Misc 9 Important

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Question 1 (i) Deleted for CBSE Board 2024 Exams

Question 1 (ii) Important Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Chapter 4 Class 11 Complex Numbers

Serial order wise

Last updated at May 29, 2023 by Teachoo

Misc 1 Evaluate: (π^18+(1/i)^25 )^3 (π^18+(1/π)^25 )^3 = (π^18+ 1/(π)^ππ )^3 = (π^18+ 1/(π Γ π^ππ ))^3 = ((π^π )^π+1/(π Γ (π^π )^ππ ))^3 Putting i2 = βπ = ((βπ)^π+1/γπ Γ (βπ)γ^12 )^3 = (βπ+1/(π Γ π))^3 = (β1+1/π)^3 Removing π from the denominator = (β1+1/πΓπ/π)^3 = (β1+π/π^π )^3 = (β1+π/((βπ)))^3 = (βπ β π )π = (β1(1+ π ))3 = (βπ)π (π + π )π = (β1)(1 + π )3 = β(π + π )π Using (a + b) 3 = a3 + b3 + 3ab(a + b) = β(13 + π3 + 3 Γ 1 Γ π (1 + π)) = β(1 + ππ +3π (1 + π)) = β(1 + ππ Γ π +3π (1 + π)) Putting i2 = βπ = β(1 +(βπ) Γ π +3π (1 + π)) = β(1 βπ +3π (1 + π)) = β(1 βπ +3π+3π Γ π) = β(1+2π+3π^π ) Putting i2 = βπ = β(1+2π+3 Γ βπ) = β(1+2πβ3) = β(2πβ2) = β2π+2 = πβππ