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AllQuestion and Answers: Page 1975

Question Number 8407 Answers: 0 Comments: 1

Question Number 8403 Answers: 3 Comments: 0

(Q.1) cos 4A=1−8cos^2 A + 8cos^2 A (Q.2) ((sec8 A −1)/(sec4 A −1)) = ((tan 8A)/(tan 2A)) (Q.3) tanA+tan(60^0 +A)+tan(120^0 +4) = 3tan 3A (Q.4) sinA sin (60^0 −A) sin(60^0 +A) = (1/4)sin3A

$$\left({Q}.\mathrm{1}\right)\:{cos}\:\mathrm{4}{A}=\mathrm{1}−\mathrm{8}{cos}^{\mathrm{2}} {A}\:+\:\mathrm{8}{cos}^{\mathrm{2}} \:{A} \\ $$$$\left({Q}.\mathrm{2}\right)\:\:\frac{{sec}\mathrm{8}\:{A}\:−\mathrm{1}}{{sec}\mathrm{4}\:{A}\:−\mathrm{1}}\:=\:\frac{{tan}\:\mathrm{8}{A}}{{tan}\:\mathrm{2}{A}} \\ $$$$\left({Q}.\mathrm{3}\right)\:\:\:{tanA}+{tan}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)+{tan}\left(\mathrm{120}^{\mathrm{0}} \right. \\ $$$$\left.+\mathrm{4}\right)\:=\:\mathrm{3}{tan}\:\mathrm{3}{A}\: \\ $$$$\left({Q}.\mathrm{4}\right)\:\:\:{sinA}\:{sin}\:\left(\mathrm{60}^{\mathrm{0}} −{A}\right)\:\:{sin}\left(\mathrm{60}^{\mathrm{0}} +{A}\right)\: \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{3}{A} \\ $$

Question Number 8402 Answers: 1 Comments: 0

Q. (cos^2 66^0 −sin^2 6^0 )(cos^2 48^0 −sin^2 12^0 ) = (1/(16))

$${Q}.\:\left({cos}^{\mathrm{2}} \mathrm{66}^{\mathrm{0}} −{sin}^{\mathrm{2}} \mathrm{6}^{\mathrm{0}} \right)\left({cos}^{\mathrm{2}} \mathrm{48}^{\mathrm{0}} −{sin}^{\mathrm{2}} \mathrm{12}^{\mathrm{0}} \right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{16}} \\ $$

Question Number 8401 Answers: 2 Comments: 0

Q .1 1+cos^2 2A=2(cos^4 A+sin^4 A) 2. sin^2 A+sin^2 (120^0 +A)+sin^2 (120^0 −A) = (3/2)

$${Q}\:.\mathrm{1}\:\:\:\mathrm{1}+{cos}^{\mathrm{2}} \mathrm{2}{A}=\mathrm{2}\left({cos}^{\mathrm{4}} \:{A}+{sin}^{\mathrm{4}} \:{A}\right) \\ $$$$\mathrm{2}.\:\:{sin}^{\mathrm{2}} {A}+{sin}^{\mathrm{2}} \left(\mathrm{120}^{\mathrm{0}} +{A}\right)+{sin}^{\mathrm{2}} \left(\mathrm{120}^{\mathrm{0}} −{A}\right) \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 8390 Answers: 1 Comments: 0

Question Number 8389 Answers: 1 Comments: 0

Question Number 8387 Answers: 1 Comments: 0

Question Number 8375 Answers: 0 Comments: 0

Evaluate : ∫_0 ^(1/2) (√(1 − x^2 )) dx By direct integration and by expanding as a power series.

$$\mathrm{Evaluate}\::\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \sqrt{\mathrm{1}\:−\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\mathrm{By}\:\mathrm{direct}\:\mathrm{integration}\:\mathrm{and}\:\mathrm{by}\:\mathrm{expanding} \\ $$$$\mathrm{as}\:\mathrm{a}\:\mathrm{power}\:\mathrm{series}. \\ $$

Question Number 8373 Answers: 1 Comments: 2

Question Number 8368 Answers: 2 Comments: 0

prove π=(6/(√3))×Σ_(x=1) ^∞ (((−1)^(x+1) )/((2x−1)×3^(x−1) ))

$$ \\ $$$${prove} \\ $$$$\pi=\frac{\mathrm{6}}{\sqrt{\mathrm{3}}}×\underset{{x}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{x}+\mathrm{1}} }{\left(\mathrm{2}{x}−\mathrm{1}\right)×\mathrm{3}^{{x}−\mathrm{1}} } \\ $$$$ \\ $$

Question Number 8367 Answers: 2 Comments: 0

Question Number 8366 Answers: 0 Comments: 0

∫(x^2 /(√(x^4 + 4))) dx

$$\int\frac{\mathrm{x}^{\mathrm{2}} }{\sqrt{\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{4}}}\:\mathrm{dx} \\ $$

Question Number 8362 Answers: 1 Comments: 0

(√(2(√(2(√(2(√(2(√(2(√(2.....))))))))))))=?

$$ \\ $$$$ \\ $$$$\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{2}.....}}}}}}=? \\ $$$$ \\ $$$$ \\ $$

Question Number 8359 Answers: 1 Comments: 1

prove that 10 divides 11^(10) −1

$${prove}\:{that}\: \\ $$$$\mathrm{10}\:{divides}\:\mathrm{11}^{\mathrm{10}} −\mathrm{1} \\ $$

Question Number 8357 Answers: 0 Comments: 1

who is bigger ? 1000^(1000) or 101^(999 ) please answer in details.

$${who}\:{is}\:{bigger}\:? \\ $$$$\mathrm{1000}^{\mathrm{1000}} \:{or}\:\mathrm{101}^{\mathrm{999}\:} \\ $$$${please}\:{answer}\:{in}\:{details}. \\ $$$$ \\ $$

Question Number 8356 Answers: 0 Comments: 0

find factor: a^(10) +a^5 +1

$${find}\:{factor}: \\ $$$${a}^{\mathrm{10}} +{a}^{\mathrm{5}} +\mathrm{1} \\ $$

Question Number 8355 Answers: 2 Comments: 2

if a+b+c=0 then (((a−b)/c)+((b−c)/a)+((c−a)/b))((a/(b−c))+(b/(c−a))+(c/(a−b)))=?

$${if}\:{a}+{b}+{c}=\mathrm{0}\:{then} \\ $$$$ \\ $$$$\left(\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}−{a}}{{b}}\right)\left(\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}\right)=? \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 8353 Answers: 1 Comments: 0

(√(240+2(√(14319))))=?

$$\sqrt{\mathrm{240}+\mathrm{2}\sqrt{\mathrm{14319}}}=? \\ $$

Question Number 8352 Answers: 1 Comments: 0

make t the subject of the formular. s = ut + (1/2)gt^2 + 3t^3

$$\mathrm{make}\:\mathrm{t}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formular}. \\ $$$$\mathrm{s}\:=\:\mathrm{ut}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gt}^{\mathrm{2}} \:+\:\mathrm{3t}^{\mathrm{3}} \\ $$

Question Number 8351 Answers: 1 Comments: 1

x^2 =816−64(√(77)) x=?[write x like a(√(c ))+b(√d])

$$ \\ $$$$\:\:\:{x}^{\mathrm{2}} =\mathrm{816}−\mathrm{64}\sqrt{\mathrm{77}} \\ $$$$\:\:\:{x}=?\left[{write}\:{x}\:{like}\:{a}\sqrt{{c}\:}+{b}\sqrt{\left.{d}\right]}\right. \\ $$$$ \\ $$

Question Number 8350 Answers: 1 Comments: 0

solve: (5/(3x+2))+(8/(4x+2))=((33)/(9x+8))

$${solve}: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{3}{x}+\mathrm{2}}+\frac{\mathrm{8}}{\mathrm{4}{x}+\mathrm{2}}=\frac{\mathrm{33}}{\mathrm{9}{x}+\mathrm{8}} \\ $$

Question Number 8347 Answers: 0 Comments: 5

a_1 =2 , a_(n+1) >a_n (a_(n+1) −a_n )^2 = 2(a_(n+1) +a_n ) a_n =?? help me please.

$${a}_{\mathrm{1}} =\mathrm{2}\:,\:\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} \\ $$$$\left({a}_{{n}+\mathrm{1}} −{a}_{{n}} \right)^{\mathrm{2}} =\:\mathrm{2}\left({a}_{{n}+\mathrm{1}} +{a}_{{n}} \right) \\ $$$$\:{a}_{{n}} =?? \\ $$$${help}\:{me}\:{please}. \\ $$

Question Number 8345 Answers: 1 Comments: 0

y=(x+2)^2 −3 Translation T_1 = ((a),(b) ) y′=x^2 a=? b=?

$${y}=\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3} \\ $$$${Translation}\:{T}_{\mathrm{1}} =\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix} \\ $$$${y}'={x}^{\mathrm{2}} \\ $$$${a}=?\:{b}=? \\ $$

Question Number 8341 Answers: 1 Comments: 0

What are necessary and sufficient conditions that (a+ib)^n is cyclic for an n not equal to 0?

$$\mathrm{What}\:\mathrm{are}\:\mathrm{necessary}\:\mathrm{and}\:\mathrm{sufficient}\:\mathrm{conditions} \\ $$$$\mathrm{that}\:\left(\mathrm{a}+\mathrm{ib}\right)^{\mathrm{n}} \:\mathrm{is}\:\mathrm{cyclic}\:\mathrm{for}\:\mathrm{an}\:\:\mathrm{n}\:\mathrm{not}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0}? \\ $$

Question Number 8338 Answers: 1 Comments: 1

Question Number 8334 Answers: 0 Comments: 0

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