Question
Without using trigonometric identity, show that:
$
\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)=0
$
$
\sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)=0
$
✓
Answer
$
\begin{aligned}
& \sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)=0 \\
& \sin \left(50^{\circ}+\theta\right)=\cos \left[90^{\circ}-\left(50^{\circ}+\theta\right)\right]=\cos \left(40^{\circ}-\theta\right) \\
& \sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right) \\
& =\cos \left(40^{\circ}-\theta\right)-\cos \left(40^{\circ}-\theta\right) \\
& =0
\end{aligned}
$
\begin{aligned}
& \sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right)=0 \\
& \sin \left(50^{\circ}+\theta\right)=\cos \left[90^{\circ}-\left(50^{\circ}+\theta\right)\right]=\cos \left(40^{\circ}-\theta\right) \\
& \sin \left(50^{\circ}+\theta\right)-\cos \left(40^{\circ}-\theta\right) \\
& =\cos \left(40^{\circ}-\theta\right)-\cos \left(40^{\circ}-\theta\right) \\
& =0
\end{aligned}
$
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