Question
If $(\sec\theta+\tan\theta)=\text{m}$ and $(\sec\theta-\tan\theta)=\text{n},$ show that mn = 1.
✓
Answer
We have $(\sec\theta+\tan\theta)=\text{m}\dots(\text{i})$
Again, $(\sec\theta-\tan\theta)=\text{n}\dots(\text{ii})$
Now, multiplying (i) and (ii), we get:
$(\sec\theta+\tan\theta)\times(\sec\theta-\tan\theta)=\text{mn}$
$\Rightarrow\sec^2\theta-\tan^2\theta=\text{mn}$
$\Rightarrow1=\text{mn}$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$\therefore\ \text{mn}=1$
Again, $(\sec\theta-\tan\theta)=\text{n}\dots(\text{ii})$
Now, multiplying (i) and (ii), we get:
$(\sec\theta+\tan\theta)\times(\sec\theta-\tan\theta)=\text{mn}$
$\Rightarrow\sec^2\theta-\tan^2\theta=\text{mn}$
$\Rightarrow1=\text{mn}$ $\big[\because\sec^2\theta-\tan^2\theta=1\big]$
$\therefore\ \text{mn}=1$
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