Question
Prove the following:$\frac{\cos(90^\circ-\text{A})\sin(90^\circ-\text{A})}{\tan(90^\circ-\text{A})}-\sin^2\text{A}=0$
✓
Answer
$\cos(90^\circ-\text{A})=\sin\text{A}$ $\tan(90^\circ-\text{A})=\cot\text{A}$
$\sin(90^\circ-\text{A})=\cos\text{A}$
$\frac{\sin\text{A}.\cos\text{A}}{\cot\text{A}}-\sin^2\text{A}=0$
$\frac{\sin\text{A}.\cos\text{A}}{\cos\text{A}}\sin\text{A}-\sin^2\text{A}$
$\sin^2\text{A}-\sin^2\text{A}=0$
$\text{L.H.S = R.H.S}$
Hence Proved.
$\sin(90^\circ-\text{A})=\cos\text{A}$
$\frac{\sin\text{A}.\cos\text{A}}{\cot\text{A}}-\sin^2\text{A}=0$
$\frac{\sin\text{A}.\cos\text{A}}{\cos\text{A}}\sin\text{A}-\sin^2\text{A}$
$\sin^2\text{A}-\sin^2\text{A}=0$
$\text{L.H.S = R.H.S}$
Hence Proved.
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