math_doubt_questionFrequently asked Questions
1.Prove1secπœƒ+tanπœƒ-1cosπœƒ=1cosπœƒ-1secπœƒ-tanπœƒLHS=1(secπœƒ+tanπœƒ)(secπœƒ-tanπœƒ)(secπœƒ-tanπœƒ)-1cosπœƒ=(secπœƒ-tanπœƒ)(sec2πœƒ-tan2πœƒ)-secπœƒ=secπœƒ-tanπœƒ-secπœƒ=secπœƒ-(tanπœƒ+secπœƒ)=1cosπœƒ-(secπœƒ+tanπœƒ)1(secπœƒ-tanπœƒ)(secπœƒ-tanπœƒ)=1cosπœƒ-(secπœƒ+tanπœƒ)1(secπœƒ-tanπœƒ)(secπœƒ-tanπœƒ)=1cosπœƒ-(sec2πœƒ-tan2πœƒ)(secπœƒ-tanπœƒ)=1cosπœƒ-(sec2πœƒ-tan2πœƒ)(secπœƒ-tanπœƒ)=1cosπœƒ-1secπœƒ-tanπœƒ=RHS
2.Prove1-secπœƒ+tanπœƒ1+secπœƒ-tanπœƒ=secπœƒ+tanπœƒ-1secπœƒ+tanπœƒ+1LHS=1-secπœƒ+tanπœƒ1+secπœƒ-tanπœƒ=(sec2πœƒ-tan2πœƒ)-(secπœƒ-tanπœƒ)(sec2πœƒ-tan2πœƒ)+(secπœƒ-tanπœƒ)=(secπœƒ-tanπœƒ)(secπœƒ+tanπœƒ)-(secπœƒ-tanπœƒ)(secπœƒ-tanπœƒ)(secπœƒ+tanπœƒ)+(secπœƒ-tanπœƒ)=(secπœƒ+tanπœƒ-1)(secπœƒ-tanπœƒ)(secπœƒ+tanπœƒ+1)(secπœƒ-tanπœƒ)=(secπœƒ+tanπœƒ-1)(secπœƒ+tanπœƒ+1)=RHS
3.Provetanπœƒ+secπœƒ-1tanπœƒ+secπœƒ+1=tanπœƒsecπœƒ+1LHS=tanπœƒ+secπœƒ-1tanπœƒ+secπœƒ+1=(tanπœƒ+secπœƒ)-(sec2πœƒ-tan2πœƒ)(tanπœƒ+secπœƒ)+(sec2πœƒ-tan2πœƒ)=(secπœƒ-tanπœƒ)(secπœƒ+tanπœƒ)-(secπœƒ-tanπœƒ)(secπœƒ-tanπœƒ)(secπœƒ+tanπœƒ)+(secπœƒ-tanπœƒ)=(secπœƒ+tanπœƒ-1)(secπœƒ-tanπœƒ)(secπœƒ+tanπœƒ+1)(secπœƒ-tanπœƒ)=(secπœƒ+tanπœƒ-1)(secπœƒ+tanπœƒ+1)=RHS
4Provetan(πœ‹4-πœƒ)=(1-tanπœƒ)(1+tanπœƒ)(usingformula:tan(A+B)=tanA+tanB1-tanAtanB)LHS=tan(πœ‹4+πœƒ)=tanπœ‹4-tanπœƒ1+tanπœ‹4tanπœƒ=1-tanπœƒ1+tanπœƒ=RHS
5Provetan5A-tan3Atan5A+tan3A=sin2Asin8ALHS=tan5A-tan3Atan5A+tan3A=sin5Acos5A-sin3Acos3Asin5Acos5A+sin3Acos3A=sin5A.cos3A-sin3A.cos5Acos3A.cos5Asin5A.cos3A+sin3A.cos5Acos3A.cos5A=sin5A.cos3A-sin3A.cos5Asin5A.cos3A+sin3A.cos5A=sin(5A-3A)sin(5A+3A)=sin2Asin8A=RHS
QuestionProve:tan50=tan40+2tan10Answertan50=tan(40+10)⟹tan50=tan40+tan101-tan40.tan10⟹tan50-tan50.tan40.tan10=tan40+tan10⟹tan50-tan50.tan(90-50).tan10=tan40+tan10⟹tan50-tan50.cot50.tan10=tan40+tan10⟹tan50-tan10=tan40+tan10⟹tan50=tan40+2tan10proved
7Proove:cotA.cot4A+1cotA.cot4A-1=cos3Acos5ALHS=cotA.cot4A+1cotA.cot4A-1=cosA.cos4AsinA.sin4A+1cosA.cos4AsinA.sin4A-1=cosA.cos4A+sinA.sin4AsinA.sin4AcosA.cos4A-sinA.sin4AsinA.sin4A=cosA.cos4A+sinA.sin4AcosA.cos4A-sinA.sin4A=cos(4A-A)cos(4A+A)=cos3Acos5A=RHS
Questionπœ” is cube root of unity, find the value of(1+πœ”)(1+πœ”2)(1+πœ”4)(1+πœ”8) AnswerAs πœ” is cube root of unity so πœ”2+πœ”+1=0 and πœ”3=1A=(1+πœ”)(1+πœ”2)(1+πœ”4)(1+πœ”8) ⟹A=(1+πœ”)(1+πœ”2)(1+πœ”4)(1+πœ”8)⟹A=(1+πœ”)(1+πœ”2)(1+πœ”πœ”3)(1+πœ”4πœ”2)⟹A=(1+πœ”)(1+πœ”2)(1+πœ”)(1+πœ”2)⟹A=(1+πœ”)(1+πœ”2)(1+πœ”)(1+πœ”2)⟹A=(1+πœ”+πœ”2-πœ”2)(1+πœ”+πœ”2-πœ”)(1+πœ”+πœ”2-πœ”2)(1+πœ”+πœ”2-πœ”)⟹A=(-πœ”2)(-πœ”)(-πœ”2)(-πœ”)⟹A=πœ”6=1
QuestionIf𝛼and𝛽arecuberootofunitythenshowthat(a)𝛼2+𝛽2+𝛼𝛽=0(b)𝛼4+𝛽4+𝛼-1𝛽-1=0AnswerCuberootofunity=xthenx3=1⟹(x3-1)=0⟹(x-1)(x2+x+1)=0Sox=1(whichisnotacomplexnumber)x2+x+1=0x=-1+i32andx=-1-i32So,𝛼=-1+i32𝛽=-1-i32 As 𝛼 and 𝛽 are cube root of unity so𝛼2+𝛼+1=0----------(a)𝛼3=1𝛽2+𝛽+1=0-----------(b)𝛽3=1𝛼+𝛽=-1(byformula,sumofroots=-ba)----(c)𝛼𝛽=1(byformula,productofroots=ca)----(d)Adding(a)and(b)𝛼2+𝛼+1+𝛽2+𝛽+1=0βŸΉπ›Ό2+𝛽2+𝛼+1+𝛽+1=0βŸΉπ›Ό2+𝛽2+𝛼+𝛽+2=0βŸΉπ›Ό2+𝛽2-1+2=0βŸΉπ›Ό2+𝛽2+1=0by(d)weknow𝛼𝛽=1So,𝛼2+𝛽2+𝛼𝛽=0[proved] (b)LHS=𝛼4+𝛽4+𝛼-1𝛽-1=𝛼4+𝛽4+2𝛼2𝛽2-2𝛼2𝛽2+1𝛼𝛽=(𝛼2+𝛽2)2-2𝛼2𝛽2+1𝛼𝛽=(-𝛼𝛽)2-2𝛼2𝛽2+1𝛼𝛽=(-1)2-2+11=1-2+1=0=RHS
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