Question
Prove the following.
secθ(1 - sinθ) (secθ + tanθ) = 1
secθ(1 - sinθ) (secθ + tanθ) = 1
✓
Answer
$\text { Taking LHS }$
$\sec \theta(1-\sin \theta)(\sec \theta+\tan \theta) $
$=\frac{1}{\cos \theta}(1-\sin \theta)\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) $
$=\frac{1}{\cos \theta}(1-\sin \theta) \frac{1}{\cos \theta}(1+\sin \theta) $
$=\frac{1}{\cos ^2 \theta}\left(1-\sin ^2 \theta\right)\left[(a+b)(a-b)=a^2-b^2\right] $
$=\frac{1}{\cos ^2 \theta}\left(\cos ^2 \theta\right)\left[\sin ^2 \theta+\cos ^2 \theta=1\right] $
$=1 $
=\text { RHS }
\text { Proved! }
$\sec \theta(1-\sin \theta)(\sec \theta+\tan \theta) $
$=\frac{1}{\cos \theta}(1-\sin \theta)\left(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}\right) $
$=\frac{1}{\cos \theta}(1-\sin \theta) \frac{1}{\cos \theta}(1+\sin \theta) $
$=\frac{1}{\cos ^2 \theta}\left(1-\sin ^2 \theta\right)\left[(a+b)(a-b)=a^2-b^2\right] $
$=\frac{1}{\cos ^2 \theta}\left(\cos ^2 \theta\right)\left[\sin ^2 \theta+\cos ^2 \theta=1\right] $
$=1 $
=\text { RHS }
\text { Proved! }
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