Question

sina +2sin2a + sin3a / cos a +2cos2a + cos 3a =tan2a ​

Answer 1

Question :- Prove that

$\rm \: \dfrac{sina + 2sin2a + sin3a}{cosa + 2cos2a + cos3a} = tan2a$

$\large\underline{\sf{Solution-}}$

Consider LHS

$\rm \: \dfrac{sina + 2sin2a + sin3a}{cosa + 2cos2a + cos3a}$

can be re-arranged as

$\rm \: = \: \dfrac{(sin3a + sina) + 2sin2a}{(cos3a + cosa) + 2cos2a}$

We know,

$\color{green}\boxed{ \rm{ \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

And

$\color{green}\boxed{ \rm{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }}$

So, using these results, we get

$\rm \: = \: \dfrac{2sin\bigg[\dfrac{3a + a}{2} \bigg]cos\bigg[\dfrac{3a - a}{2} \bigg] + 2sin2a}{2cos\bigg[\dfrac{3a + a}{2} \bigg]cos\bigg[\dfrac{3a - a}{2} \bigg] + 2cos2a}$

$\rm \: = \: \dfrac{2sin2a \: cosa \: + \: 2sin2a}{2cos2a \: cosa \: + \: 2cos2a}$

$\rm \: = \: \dfrac{2sin2a \:( cosa \: + \: 1)}{2cos2a \:( cosa \: + \: 1)}$

$\rm \: = \: \dfrac{sin2a}{cos2a}$

$\rm \: = \: tan2a$

Hence,

$\color{green}\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{sina + 2sin2a + sin3a}{cosa + 2cos2a + cos3a} = tan2a \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(x - y) = sinx \: cosy \: - \: siny \: cosx}\\ \\ \bigstar \: \bf{sin(x + y) = sinx \: cosy \: + \: siny \: cosx}\\ \\ \bigstar \: \bf{cos(x + y) = cosx \: cosy \: - \: sinx \: siny}\\ \\ \bigstar \: \bf{cos(x - y) = cosx \: cosy \:+\: siny \: sinx}\\ \\ \bigstar \: \bf{tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany} }\\ \\ \bigstar \: \bf{tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

Question :- Prove that

\begin{gathered}\rm \: \dfrac{sina + 2sin2a + sin3a}{cosa + 2cos2a + cos3a} = tan2a \\ \end{gathered}

cosa+2cos2a+cos3a

sina+2sin2a+sin3a

=tan2a

\large\underline{\sf{Solution-}}

Solution−

Consider LHS

\begin{gathered}\rm \: \dfrac{sina + 2sin2a + sin3a}{cosa + 2cos2a + cos3a} \\ \end{gathered}

cosa+2cos2a+cos3a

sina+2sin2a+sin3a

can be re-arranged as

\begin{gathered}\rm \: = \: \dfrac{(sin3a + sina) + 2sin2a}{(cos3a + cosa) + 2cos2a} \\ \end{gathered}

=

(cos3a+cosa)+2cos2a

(sin3a+sina)+2sin2a

We know,

\begin{gathered}\color{green}\boxed{ \rm{ \:sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }} \\ \end{gathered}

sinx+siny=2sin[

2

x+y

]cos[

2

x−y

]

And

\begin{gathered}\color{green}\boxed{ \rm{ \:cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg] \: }} \\ \end{gathered}

cosx+cosy=2cos[

2

x+y

]cos[

2

x−y

]

So, using these results, we get

\begin{gathered}\rm \: = \: \dfrac{2sin\bigg[\dfrac{3a + a}{2} \bigg]cos\bigg[\dfrac{3a - a}{2} \bigg] + 2sin2a}{2cos\bigg[\dfrac{3a + a}{2} \bigg]cos\bigg[\dfrac{3a - a}{2} \bigg] + 2cos2a} \\ \end{gathered}

=

2cos[

2

3a+a

]cos[

2

3a−a

]+2cos2a

2sin[

2

3a+a

]cos[

2

3a−a

]+2sin2a

\begin{gathered}\rm \: = \: \dfrac{2sin2a \: cosa \: + \: 2sin2a}{2cos2a \: cosa \: + \: 2cos2a} \\ \end{gathered}

=

2cos2acosa+2cos2a

2sin2acosa+2sin2a

\begin{gathered}\rm \: = \: \dfrac{2sin2a \:( cosa \: + \: 1)}{2cos2a \:( cosa \: + \: 1)} \\ \end{gathered}

=

2cos2a(cosa+1)

2sin2a(cosa+1)

\begin{gathered}\rm \: = \: \dfrac{sin2a}{cos2a} \\ \end{gathered}

=

cos2a

sin2a

\begin{gathered}\rm \: = \: tan2a \\ \end{gathered}

=tan2a

Hence,

\begin{gathered}\color{green}\rm\implies \:\boxed{ \rm{ \:\rm \: \dfrac{sina + 2sin2a + sin3a}{cosa + 2cos2a + cos3a} = tan2a \: }} \\ \end{gathered}

⟹

cosa+2cos2a+cos3a

sina+2sin2a+sin3a

=tan2a

\rule{190pt}{2pt}

Additional Information :-

\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{sin(x - y) = sinx \: cosy \: - \: siny \: cosx}\\ \\ \bigstar \: \bf{sin(x + y) = sinx \: cosy \: + \: siny \: cosx}\\ \\ \bigstar \: \bf{cos(x + y) = cosx \: cosy \: - \: sinx \: siny}\\ \\ \bigstar \: \bf{cos(x - y) = cosx \: cosy \:+\: siny \: sinx}\\ \\ \bigstar \: \bf{tan(x + y) = \dfrac{tanx + tany}{1 - tanx \: tany} }\\ \\ \bigstar \: \bf{tan(x - y) = \dfrac{tanx - tany}{1 + tanx \: tany} }\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

MoreFormulae

MoreFormulae

★sin(x−y)=sinxcosy−sinycosx

★sin(x+y)=sinxcosy+sinycosx

★cos(x+y)=cosxcosy−sinxsiny

★cos(x−y)=cosxcosy+sinysinx

★tan(x+y)=

1−tanxtany

tanx+tany

★tan(x−y)=

1+tanxtany

tanx−tany

Hope it helps you

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