Question

Cos10-sin10/cos10+sin10=cot55

Answer 1

Question :-

Prove that

$\rm \: \frac{cos10\degree - sin10\degree }{cos10\degree + sin10\degree } = cot55\degree$

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

Consider LHS

$\rm \: \frac{cos10\degree - sin10\degree }{cos10\degree + sin10\degree }$

Divide numerator and denominator by cos10°, we get

$\rm \: = \: \frac{ \dfrac{cos10\degree }{cos10\degree } - \dfrac{sin10\degree }{cos10\degree } }{\dfrac{cos10\degree }{cos10\degree } + \dfrac{sin10\degree }{cos10\degree }}$

$\rm \: = \: \frac{1 - tan10\degree }{1 + tan10\degree }$

can be further rewritten as

$\rm \: = \: \frac{1 - tan10\degree }{1 + 1 \times tan10\degree }$

can be further rewritten as

$\rm \: = \: \frac{tan45\degree - tan10\degree }{1 + tan45\degree \times tan10\degree }$

We know,

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

So, using this result, we get

$\rm \: = \:tan(45\degree - 10\degree )$

$\rm \: = \:tan35\degree$

$\rm \: = \:tan(90\degree - 55\degree )$

$\rm \: = \:cot55\degree$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \frac{cos10\degree - sin10\degree }{cos10\degree + sin10\degree } = cot55\degree \: \: }}$

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Additional Information :-

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

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

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

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

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

Answer 2

Step-by-step explanation:

R.H.S $\pmb{\tt =\cot 55^{\circ}=\cot \left(45^{\circ}+10^{\circ}\right) }$

$\color{olive} \pmb{\[ \begin{array}{l} \tt =\dfrac{\cot 45^{\circ} \cot 10^{\circ}-1}{\cot 10^{\circ}+\cot 45^{\circ}}=\dfrac{\cot 10^{\circ}-1}{\cot 10^{\circ}+1} \\ \\ \tt=\dfrac{\dfrac{\cos 10^{\circ}}{\sin 10^{\circ}}-1}{\dfrac{\cos 10^{\circ}}{\sin 10^{\circ}}+1} \: \: =\dfrac{\dfrac{\cos 10^{\circ}-\sin 10^{\circ}}{\sin 10^{\circ}}}{\dfrac{\cos 10^{\circ}+\sin 10^{\circ}}{\sin 10^{\circ}}} \\ \\ \tt=\dfrac{\cos 10^{\circ}-\sin 10^{\circ}}{\cos 10^{\circ}+\sin 10^{\circ}}=\text { L.H.S., Hence proved } \end{array} \]}$