Math, asked by mrshubhamyt94, 2 months ago

sin3x/cosx + cos3x/sin x = 2Cot 2x​

Answers

Answered by mathdude500
4

Question :-

Prove that

\rm \: \dfrac{sin3x}{cosx} + \dfrac{cos3x}{sinx} = 2 \: cot2x \\

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

Consider, LHS

\rm \: \dfrac{sin3x}{cosx} + \dfrac{cos3x}{sinx} \\

On taking LCM, we get

\rm \:=  \:  \dfrac{sin3x \: sinx \:  +  \: cos3x \: cosx}{cosx \: sinx}  \\

can be re-arranged as

\rm \:=  \:  \dfrac{cos3x \: cosx \:  +  \: sin3x \: sinx}{cosx \: sinx}  \\

We know,

\boxed{\sf{  \:\rm \: cosxcosy + sinxsiny = cos(x - y) \: }} \\

So, using this result, we get

\rm \: =  \: \dfrac{cos(3x - x)}{sinx \: cosx}  \\

\rm \: =  \: \dfrac{cos2x}{sinx \: cosx}  \\

\rm \: =  \: \dfrac{2cos2x}{2 \: sinx \: cosx}  \\

\rm \: =  \: \dfrac{2cos2x}{sin2x}  \\

\rm \: =  \: 2 \: cot2x \\

Hence,

\rm\implies \:\boxed{\sf{  \: \: \rm \: \dfrac{sin3x}{cosx} + \dfrac{cos3x}{sinx} = 2 \: cot2x  \:  \: }}\\

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Additional Information :-

\boxed{\sf{  \:\rm \: cosxcosy -  sinxsiny = cos(x  +  y) \: }} \\

\boxed{\sf{  \:\rm \: sin(x + y) = sinxcosy + sinycosx \:  \: }} \\

\boxed{\sf{  \:\rm \: sin(x -  y) = sinxcosy  -  sinycosx \:  \: }} \\

\boxed{\sf{  \:\rm \: tan(x + y) =  \frac{tanx + tany}{1 - tanx \: tany}  \:  \: }} \\

\boxed{\sf{  \:\rm \: tan(x - y) =  \frac{tanx -  tany}{1 + tanx \: tany}  \:  \: }} \\

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

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

\boxed{\sf{  \:\rm \: sin2x = 2sinxcosx =  \frac{2tanx}{1 +  {tan}^{2} x}  \:  \: }} \\

\boxed{\sf{  \:\rm \: cos2x =  {cos}^{2}x -  {sin}^{2}x = 1 -  {2sin}^{2}x =  {2cos}^{2}x - 1 =  \frac{1 -  {tan}^{2}x}{1 +  {tan}^{2} x}  \: }} \\

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