Question
Prove the following trigonometric identities.
$(1+\cot\text{A}-\text{cosec A})(1+\tan\text{A}+\sec\text{A})=2$
$(1+\cot\text{A}-\text{cosec A})(1+\tan\text{A}+\sec\text{A})=2$
✓
Answer
$\text{L.H.S}=(1+\cot\text{A}-\text{cosec A})(1+\tan\text{A}+\sec\text{A})$
$=\Big(1+\frac{\cos\text{A}}{\sin\text{A}}-\frac{1}{\sin\text{A}}\Big)\Big(1+\frac{\sin\text{A}}{\cos\text{A}}+\frac{1}{\cos\text{A}}\Big)$
$=\Big(\frac{\sin\text{A}+\cos\text{A}-1}{\sin\text{A}}\Big)\Big(\frac{\cos\text{A}+\sin\text{A}+1}{\cos\text{A}}\Big)$
$=\frac{(\sin\text{A}+\cos\text{A})^2-1}{\sin\text{A}\cos\text{A}}$
$=\frac{1+2\sin\text{A}\cos\text{A}-1}{\sin\text{A}\cos\text{A}} \big[\because \sin^2\text{A}+\cos^2\text{A}=1\big]$
$=2=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
$=\Big(1+\frac{\cos\text{A}}{\sin\text{A}}-\frac{1}{\sin\text{A}}\Big)\Big(1+\frac{\sin\text{A}}{\cos\text{A}}+\frac{1}{\cos\text{A}}\Big)$
$=\Big(\frac{\sin\text{A}+\cos\text{A}-1}{\sin\text{A}}\Big)\Big(\frac{\cos\text{A}+\sin\text{A}+1}{\cos\text{A}}\Big)$
$=\frac{(\sin\text{A}+\cos\text{A})^2-1}{\sin\text{A}\cos\text{A}}$
$=\frac{1+2\sin\text{A}\cos\text{A}-1}{\sin\text{A}\cos\text{A}} \big[\because \sin^2\text{A}+\cos^2\text{A}=1\big]$
$=2=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
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