what are the demerits of an election competion?
Answers
Explanation:
\large\underline{\sf{Solution-}}
Solution−
We have with us to find the value of
\red{\rm :\longmapsto\:\sf\dfrac{cos^3\beta}{cos\alpha}+\dfrac{sin^3\beta}{sin\alpha}}:⟼
cosα
cos
3
β
+
sinα
sin
3
β
can be rewritten as
\rm \: = \:\sf\dfrac{4cos^3\beta}{4cos\alpha}+\dfrac{4sin^3\beta}{4sin\alpha}=
4cosα
4cos
3
β
+
4sinα
4sin
3
β
We know,
\begin{gathered}\red{\rm :\longmapsto\:cos3x = {4cos}^{3}x - 3cosx} \\ \red{\rm :\longmapsto\:it \: means} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \red{\rm :\longmapsto\: {4cos}^{3}x = cos3x + 3cosx}\end{gathered}
:⟼cos3x=4cos
3
x−3cosx
:⟼itmeans
:⟼4cos
3
x=cos3x+3cosx
Also,
\red{\rm :\longmapsto\: {4sin}^{3}x = 3sinx - sin3x}:⟼4sin
3
x=3sinx−sin3x
So, using this, we get
\rm \: = \:\sf\dfrac{cos3 \beta + 3cos \beta }{4cos\alpha}+\dfrac{3sin \beta - sin3 \beta }{4sin\alpha}=
4cosα
cos3β+3cosβ
+
4sinα
3sinβ−sin3β
\rm \: = \:\dfrac{cos3 \beta sin \alpha + 3cos \beta sin \alpha + 3sin \beta cos \alpha - sin3 \beta cos \alpha }{4sin \alpha cos \alpha }=
4sinαcosα
cos3βsinα+3cosβsinα+3sinβcosα−sin3βcosα
\rm \: = \:\dfrac{cos3 \beta sin \alpha- sin3 \beta cos \alpha + 3(cos \beta sin \alpha + sin \beta cos \alpha)}{2(2sin \alpha cos \alpha) }=
2(2sinαcosα)
cos3βsinα−sin3βcosα+3(cosβsinα+sinβcosα)
We know,
\boxed{ \bf{ \: sin(x + y) = sinxcosy + sinycosx}}
sin(x+y)=sinxcosy+sinycosx
and
\boxed{ \bf{ \: sin(x - y) = sinxcosy - sinycosx}}
sin(x−y)=sinxcosy−sinycosx
and
\boxed{ \bf{ \: sin2x = 2sinx \: cosx}}
sin2x=2sinxcosx
So, using these,
\rm \: = \:\dfrac{sin( - 3\beta + \alpha ) + 3sin( \alpha + \beta )}{2sin2 \alpha }=
2sin2α
sin(−3β+α)+3sin(α+β)
\rm \: = \:\dfrac{sin( \alpha - 3\beta ) - sin( \alpha + \beta ) + sin( \alpha + \beta )+ 3sin( \alpha + \beta )}{2sin2 \alpha }=
2sin2α
sin(α−3β)−sin(α+β)+sin(α+β)+3sin(α+β)
\rm \: = \:\dfrac{2cos\bigg[\dfrac{ \alpha - 3\beta + \alpha + \beta }{2} \bigg]sin\bigg[\dfrac{\alpha - 3\beta - \alpha - \beta }{2} \bigg] + 4sin( \alpha + \beta )}{2sin2 \alpha }=
2sin2α
2cos[
2
α−3β+α+β
]sin[
2
α−3β−α−β
]+4sin(α+β)
\rm \: = \:\dfrac{2cos( \alpha - \beta )sin( - 2 \beta ) + 4sin( \alpha + \beta )}{2sin2 \alpha }=
2sin2α
2cos(α−β)sin(−2β)+4sin(α+β)
\purple{\bf \: = \:\dfrac{ - \: 2cos( \alpha - \beta )sin 2 \beta + 4sin( \alpha + \beta )}{2sin2 \alpha } - - - (1)}=
2sin2α
−2cos(α−β)sin2β+4sin(α+β)
−−−(1)
Now, Consider
\rm :\longmapsto\:\sf\dfrac{cos\alpha}{cos\beta}+\dfrac{sin\alpha}{sin\beta}= -1:⟼
cosβ
cosα
+
sinβ
sinα
=−1
\rm :\longmapsto\:\dfrac{cos \alpha sin \beta + sin \alpha cos \beta }{sin \beta cos \beta } = - 1:⟼
sinβcosβ
cosαsinβ+sinαcosβ
=−1
\rm :\longmapsto\:\dfrac{sin( \alpha + \beta) }{sin \beta cos \beta } = - 1:⟼
sinβcosβ
sin(α+β)
=−1
\rm :\longmapsto\:sin( \alpha + \beta ) = - sin \beta cos \beta:⟼sin(α+β)=−sinβcosβ
\rm :\longmapsto\:2sin( \alpha + \beta ) = - 2 sin \beta cos \beta:⟼2sin(α+β)=−2sinβcosβ
\rm :\longmapsto\:2sin( \alpha + \beta ) = - sin2 \beta:⟼2sin(α+β)=−sin2β
So, equation (1), can be rewritten as
\rm \: = \:\dfrac{ 4\: cos( \alpha - \beta )sin( \alpha + \beta ) + 4sin( \alpha + \beta )}{2sin2 \alpha }=
2sin2α
4cos(α−β)sin(α+β)+4sin(α+β)
\rm \: = \:\dfrac{ (\: cos( \alpha - \beta ) + 1)4sin( \alpha + \beta )}{2sin2 \alpha }=
2sin2α
(cos(α−β)+1)4sin(α+β)
\purple{\bf \: = \:\dfrac{ 2(\: cos( \alpha - \beta ) + 1)sin( \alpha + \beta )}{sin2 \alpha } - - (2)}=
sin2α
2(cos(α−β)+1)sin(α+β)
−−(2)
Now, we know that
\rm :\longmapsto\:sin2 \alpha + sin2 \beta = 2sin( \alpha + \beta )cos( \alpha - \beta ):⟼sin2α+sin2β=2sin(α+β)cos(α−β)
\rm :\longmapsto\:sin2 \alpha + sin2 \beta = - sin2 \beta cos( \alpha - \beta ):⟼sin2α+sin2β=−sin2βcos(α−β)
\red{\bigg \{ \because \: - sin2 \beta = 2sin( \alpha + \beta ) \bigg \}}{∵−sin2β=2sin(α+β)}
\rm :\longmapsto\:sin2 \alpha = - sin2 \beta - sin2 \beta cos( \alpha - \beta ):⟼sin2α=−sin2β−sin2βcos(α−β)
\rm :\longmapsto\:sin2 \alpha = - sin2 \beta (1 + cos( \alpha - \beta )):⟼sin2α=−sin2β(1+cos(α−β))
\bf :\longmapsto\:sin2 \alpha = 2 sin( \alpha + \beta) (1 + cos( \alpha - \beta )) - - - (3):⟼sin2α=2sin(α+β)(1+cos(α−β))−−−(3)
\red{\bigg \{ \because \: - sin2 \beta = 2sin( \alpha + \beta ) \bigg \}}{∵−sin2β=2sin(α+β)}
On substituting equation (3) in (2), we get
\rm \: = \:\dfrac{ sin2 \alpha }{sin2 \alpha }=
sin2α
sin2α
\rm \: = \:1=1
Hence,
\red{\bf :\longmapsto\:\bf\dfrac{cos^3\beta}{cos\alpha}+\dfrac{sin^3\beta}{sin\alpha} = 1}:⟼
cosα
cos
3
β
+
sinα
sin
3
β
=1
Answer:
The demerits of an electoral competition are given below:
(i) It creates disunity and factionalism in every locality. People often complain of party-politics.
(ii) Different political parties and leaders often level allegations against one another. Parties and candidates often use dirty tricks to win elections.
(iii) It is often said that the pressure to win electoral