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