Question
Prove that:
$\text{cosec}(67^\circ+\theta)-\sec(23^\circ-\theta)=0$
$\text{cosec}(67^\circ+\theta)-\sec(23^\circ-\theta)=0$
✓
Answer
$\text{L.H.S.}=\text{cosec}(67^\circ+\theta)-\sec(23^\circ-\theta)$
$=\text{cosec}\big\{90^\circ-(23^\circ-\theta)\big\}-\sec(23^\circ-\theta)$
$=\sec(23^\circ+\theta)-\sec(23^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
$=\text{cosec}\big\{90^\circ-(23^\circ-\theta)\big\}-\sec(23^\circ-\theta)$
$=\sec(23^\circ+\theta)-\sec(23^\circ-\theta)$
$=0$
$=\text{R.H.S.}$
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