The value of \( \dfrac{\sin70^\circ+\cos40^\circ}{\cos70^\circ+\sin40^\circ} \) is
Solution:
Using,
\[
\cos\theta=\sin(90^\circ-\theta)
\]
\[
\cos40^\circ=\sin50^\circ
\]
and
\[
\cos70^\circ=\sin20^\circ
\]
Therefore,
\[
=
\frac{\sin70^\circ+\sin50^\circ}
{\sin20^\circ+\sin40^\circ}
\]
Using identity,
\[
\sin A+\sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
=
\frac{
2\sin60^\circ\cos10^\circ
}
{
2\sin30^\circ\cos10^\circ
}
\]
\[
=
\frac{
2\left(\frac{\sqrt3}{2}\right)\cos10^\circ
}
{
2\left(\frac12\right)\cos10^\circ
}
\]
\[
=\sqrt3
\]
\[
\boxed{\sqrt3}
\]