if sina+coseca=2then find sina+coseca,forn€N
Answers
Solving
sin
x
+
1
sin
x
=
2
→
sin
2
x
−
2
sin
x
+
1
=
0
we have
sin
x
=
1
then
sin
n
x
+
1
sin
n
x
=
2
This can be also demonstrated by finite induction.
1) First it is true for
n
=
1
(with sin x = 1 of course)
2) Supposing that it is true for
n
then
3) Prove that it is true for
n
+
1
It is quite easy and it is left as an exercise.
Answer:
Answer :- 7
Given that
\sf \: \alpha \: and \: \beta \: are \: the \: zeroes \: of \: {x}^{2} - 6x + yαandβarethezeroesofx
2
−6x+y
We know that,
\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}
Sum of the zeroes=
coefficient of x
2
−coefficient of x
\bf\implies \: \alpha + \beta = - \dfrac{( - 6)}{1} = 6 - - - (1)⟹α+β=−
1
(−6)
=6−−−(1)
Also,
Given that,
\rm :\longmapsto\:3 \alpha + 2 \beta = 20:⟼3α+2β=20
\rm :\longmapsto\: \alpha + 2 \alpha + 2 \beta = 20:⟼α+2α+2β=20
\rm :\longmapsto\: \alpha + 2( \alpha + \beta) = 20:⟼α+2(α+β)=20
\rm :\longmapsto\: \alpha + 2 \times 6 = 20:⟼α+2×6=20
\rm :\longmapsto\: \alpha + 12 = 20:⟼α+12=20
\rm :\longmapsto\: \alpha = 20 - 1:⟼α=20−1
\bf\implies \: \alpha = 8⟹α=8
\sf \: Since, \: \alpha \:is \: the \: zeroes \: of \: {x}^{2} - 6x + ySince,αisthezeroesofx
2
−6x+y
\sf \: \therefore \: \alpha \: = 8 \: must \: satisfy \: {x}^{2} - 6x + y∴α=8mustsatisfyx
2
−6x+y
\rm :\implies\: {8}^{2} - 6 \times 8 + y = 0:⟹8
2
−6×8+y=0
\rm :\implies\: 64 - 48 + y = 0:⟹64−48+y=0
\rm :\implies\: 16+ y = 0:⟹16+y=0
\rm :\implies\: y = - \: 16:⟹y=−16
Answer :- 8
Given that
\sf \: a - b,a,a + b \: are \: zeroes \: of \: {x}^{3} - {3x}^{2} + x + 1a−b,a,a+barezeroesofx
3
−3x
2
+x+1
We know that,
\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{3}}}}
Sum of the zeroes=
coefficient of x
3
−coefficient of x
\rm :\implies\:a - b + a + a + b = - \dfrac{( - 3)}{1}:⟹a−b+a+a+b=−
1
(−3)
\rm :\implies\:3a= 3:⟹3a=3
\pink{\bf\implies \:a = 1}⟹a=1
Also,
We know that,
\boxed{\red{\sf Product\ of\ the\ zeroes= - \: \frac{Constant}{coefficient\ of\ x^{3}}}}
Product of the zeroes=−
coefficient of x
3
Constant
\rm :\implies\:(a - b)(a)(a + b) = - \dfrac{(1)}{1}:⟹(a−b)(a)(a+b)=−
1
(1)
Put a = 1, we get
\rm :\longmapsto\:(1 - b)(1)(1 + b) = - 1:⟼(1−b)(1)(1+b)=−1
\rm :\longmapsto\:1 - {b}^{2} = - 1:⟼1−b
2
=−1
\rm :\longmapsto\:{b}^{2} = 2:⟼b
2
=2
\purple{\bf\implies \:b = \pm \: \sqrt{2} }⟹b=±
2
Answer :- 9
Let
'S' represents the Sum of the zeroes of a quadratic polynomial.
and
Let
'P' represents the Product of the zeroes of a quadratic polynomial.
Then,
Required Quadratic polynomial is
\: \: \: \: \: \: \: \: \purple{ \boxed{ \bf \:f(x) = k( {x}^{2} - Sx + P) \: where \: k \ne \: 0}}
f(x)=k(x
2
−Sx+P)wherek
=0