Question
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
✓
Answer
$
\begin{aligned}
& (1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta) \\
& =\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right) \\
& =\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right) \\
& =\frac{(\sin \theta+\cos \theta)^2-(1)^2}{\sin \theta \cos \theta} \\
& =\frac{\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\
& =\frac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\
& =\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2
\end{aligned}
$
\begin{aligned}
& (1+\cot \theta-\operatorname{cosec} \theta)(1+\tan \theta+\sec \theta) \\
& =\left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right) \\
& =\left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right) \\
& =\frac{(\sin \theta+\cos \theta)^2-(1)^2}{\sin \theta \cos \theta} \\
& =\frac{\sin ^2 \theta+\cos ^2 \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\
& =\frac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta} \\
& =\frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2
\end{aligned}
$
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