Question
Prove that:
If $\tan \theta+\frac{1}{\tan \theta}=2$, then show that $\tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2$
If $\tan \theta+\frac{1}{\tan \theta}=2$, then show that $\tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2$
✓
Answer
Given,
$\left(\tan \theta+\frac{1}{\tan \theta}\right)=2$
Squaring both side,
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2 \tan \theta\left(\frac{1}{\tan \theta}\right)=4\left[(a+b)^2=a^2+b^2+2 a b\right]$
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2=4$
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2$
Hence, Proved!
$\left(\tan \theta+\frac{1}{\tan \theta}\right)=2$
Squaring both side,
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2 \tan \theta\left(\frac{1}{\tan \theta}\right)=4\left[(a+b)^2=a^2+b^2+2 a b\right]$
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}+2=4$
$\Rightarrow \tan ^2 \theta+\frac{1}{\tan ^2 \theta}=2$
Hence, Proved!
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