Math, asked by murali1767, 21 days ago

If a+b+c=135,prove that(1+cotA)(1+cotB)=2

Answers

Answered by poojagchavan24
1

Answer:

Given A+B=135

⟹cot(A+B)=cot135

cotA+cotB

cotAcotB−1

=−1

⟹cotA+cotB+cotAcotB=+1

⟹cotA+cotB+cotAcotB+1=2

⟹cotA(1+cotB)+(1+cotB)=2

⟹(1+cotA)(1+cotB)=2

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Answered by mathdude500
6

Appropriate Question :-

If A + B = 135°, prove that (1 + cotA) (1 + cotB) = 2

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

Given that,

\rm \: A + B = 135 \degree \\

So,

\rm\implies \:cot(A + B) = cot135 \degree \\

\rm \:  \frac{cotA \: cotB \:  -  \: 1}{cotB + cotA}  = cot(90 \degree + 45\degree ) \\

\rm \:  \frac{cotA \: cotB \:  -  \: 1}{cotB + cotA}  =  - tan 45\degree  \\

\rm \:  \frac{cotA \: cotB \:  -  \: 1}{cotB + cotA}  =  - 1 \\

\rm \: cotA \: cotB \:  -  \: 1 \:  =  \:  - cotA - cotB \\

\rm \: cotA \: cotB +  cotA  + cotB = 1 \\

On adding 1 on both sides, we get

\rm \: cotA \: cotB +  cotA  + cotB  + 1= 1  + 1\\

\rm \: cotB(cotA + 1) + 1(cotA + 1) = 2 \\

\rm \: (cotA + 1)(cotB + 1) = 2 \\

Hence,

\color{blue}\rm\implies \:\boxed{ \rm{ \:(1 + cotA) \: (1 + cotB) = 2}} \\

\rule{190pt}{2pt}

Formulae Used :-

\color{green}\boxed{ \rm{ \:cot(90\degree  + x) =  -  \: tanx \: }} \\

\color{green}\boxed{ \rm{ \:cot(x + y) =  \frac{cotx \: coty \:  -  \: 1}{cotx + coty} \: }} \\

\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}

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