Question

sin70-cos40/cos50-sin20=1/√3

Answer 1

=sin70-sin50/cos50-cos70
=2cos(70+50)sin(70-50)/2sin(70+50)sin(70-50)
=2cos120°sin20°/2sin120°sin20°
=2*(-1/2)/2(√3/2)
=-1/√3
I getting the answer I'm minus sign,please recheck the question or may be the question is wrong

Answer 2

Given: Here we have the given equation sin70-cos40/cos50-sin20=1/√3

To solve: Here we have to solve the given equation

Solution:

Here we have the given equation

$\[\begin{align} & =\frac{sin70-sin50}{cos50-cos70} \\ & =\frac{2cos\left( 70+50 \right)sin\left( 70-50 \right)}{2sin\left( 70+50 \right)sin\left( 70-50 \right)} \\ & =\frac{2cos120{}^\circ sin20{}^\circ }{2sin120{}^\circ sin20{}^\circ } \\ & =\frac{\cos \left( 90+30 \right)}{\sin \left( 90+30 \right)} \\ & =\frac{-\sin 30}{\cos 30} \\ & =\frac{\left( -1/2 \right)}{\left( \surd 3/2 \right)} \\ & =-1/\surd 3 \\ \end{align}\]$$\[\begin{align} & =\frac{sin70-sin50}{cos50-cos70} \\ & =\frac{2cos\left( 70+50 \right)sin\left( 70-50 \right)}{2sin\left( 70+50 \right)sin\left( 70-50 \right)} \\ & =\frac{2cos{{120}^{{}^\circ }}sin{{20}^{{}^\circ }}}{2sin{{120}^{{}^\circ }}sin{{20}^{{}^\circ }}} \\ & \\ \end{align}\]$$\[=\frac{sin70-sin50}{cos50-cos70}\]$

$\[=\frac{2cos\left( 70+50 \right)sin\left( 70-50 \right)}{2sin\left( 70+50 \right)sin\left( 70-50 \right)}\]$

$\[=\frac{2cos{{120}^{{}^\circ }}sin{{20}^{{}^\circ }}}{2sin{{120}^{{}^\circ }}sin{{20}^{{}^\circ }}}\]$

$& =\frac{sin70-sin50}{cos50-cos70}$

$& =\frac{2cos\left( 70+50 \right)sin\left( 70-50 \right)}{2sin\left( 70+50 \right)sin\left( 70-50 \right)} \\ & =\frac{2cos120{}^\circ sin20{}^\circ }{2sin120{}^\circ sin20{}^\circ } \\ & =\frac{\cos \left( 90+30 \right)}{\sin \left( 90+30 \right)} \\ & =\frac{-\sin 30}{\cos 30} \\ & =\frac{\left( -1/2 \right)}{\left( \surd 3/2 \right)} \\ & =-1/\surd 3$

Final answer:

Hence the answer is $-1/\sqrt3$