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Prove the following trigonometric identities:
\( \sec A(1-\sin A)(\sec A+\tan A)=1 \)
To do:
We have to prove that \( \sec A(1-\sin A)(\sec A+\tan A)=1 \).
Solution:
We know that,
$\tan A=\frac{\sin A}{\cos A}=\sin A\sec A$.....(i)
$\sec^2 A-tan ^{2} A=1$.......(ii)
Therefore,
$\sec A(1-\sin A)(\sec A+\tan A)=(\sec A-\sec A\sin A)(\sec A+\tan A)$
$=(\sec A-\tan A)(\sec A+\tan A)$ (From (i))
$=\sec^2 A-tan ^{2} A$ [$(a+b)(a-b)=a^2-b^2$]
$=1$ (From (ii))
Hence proved.
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