Question

Show $\left|\begin{array}{cc}\sin \left(10°\right)& -\cos \left(10°\right)\\ \sin \left(80°\right)& \cos \left(80°\right)\end{array}\right|=1$

Answer 1

Solve for the required proof:

Consider the L.H.S

$\left|\begin{array}{cc}\sin \left(10°\right)& -\cos \left(10°\right)\\ \sin \left(80°\right)& \cos \left(80°\right)\end{array}\right|=\sin \left(10°\right)\cos \left(80°\right)-\left(-\cos \left(10°\right)\sin \left(80°\right)\right)\left[\because \left|\begin{array}{cc}a& b\\ c& d\end{array}\right|=ad-bc\right]$

$$\begin{gathered} =\sin \left(10°\right)\cos \left(80°\right)+\cos \left(10°\right)\sin \left(80°\right) \\ =\sin \left(10°+80°\right)\left[\because \sin \left(A+B\right)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)\right] \\ =\sin \left(90°\right) \\ =1 \end{gathered}$$

$\Rightarrow$ L.H.S$=$ R.H.S

Hence,it is proved that $\left|\begin{array}{cc}\sin \left(10°\right)& -\cos \left(10°\right)\\ \sin \left(80°\right)& \cos \left(80°\right)\end{array}\right|\mathbf{=}\mathbf{1}$.