Question
If $\sec\theta+\tan\theta=\text{p},$ prove that:$\tan\theta=\frac12\Big(\text{p}+\frac{1}{\text{p}}\Big)$
✓
Answer
We have,$\sec\theta+\tan\theta=\text{p}\dots(1)$
$\Rightarrow\frac{\sec\theta+\tan\theta}{1}\times\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{1}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{p}}\dots(2)$
Subtracting (1) and (2), we get:
$2\tan\theta=\Big(\text{p}-\frac{1}{\text{p}}\Big)$
$\Rightarrow\tan\theta=\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)$
$\Rightarrow\frac{\sec\theta+\tan\theta}{1}\times\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\frac{1}{\sec\theta-\tan\theta}=\text{p}$
$\Rightarrow\sec\theta-\tan\theta=\frac{1}{\text{p}}\dots(2)$
Subtracting (1) and (2), we get:
$2\tan\theta=\Big(\text{p}-\frac{1}{\text{p}}\Big)$
$\Rightarrow\tan\theta=\frac12\Big(\text{p}-\frac{1}{\text{p}}\Big)$
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