Question
If $\tan\theta+\sec\theta=\text{l},$ then prove that $\sec\theta=\frac{\text{l}^2+1}{2\text{l}}.$
✓
Answer
Recall identity $\sec^2\theta-\tan^2\theta=1$ and now change $\sec\theta+\tan\theta\text{ to }\sec^2\theta-\tan^2\theta$by multiplying and dividing the given expression to $(\sec\theta-\tan\theta).$
$\sec\theta+\tan\theta=\text{l}\ \ [\text{Given}]\ (\text{I})$
$\Rightarrow\ (\sec\theta+\tan\theta)\frac{(\sec\theta-\tan\theta)}{\sec\theta-\tan\theta}=\text{l}$
$\Rightarrow\ \frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{l}\ [\because1+\tan^2\theta=\sec^2\theta]$
$\Rightarrow\ \frac{1}{\sec\theta-\tan\theta}=\text{l}$
$\text{or}\ \sec\theta-\tan\theta=\frac{1}{\text{l}}\ \ (\text{II})$
Now, get $\sec\theta$ by eliminating $\tan\theta$ from (I) and (II).
It can be obtained by adding (I) and (II).
$\Rightarrow\ 2\sec\theta=\text{l}+\frac{1}{\text{l}}$
$\Rightarrow\ 2\sec\theta=\frac{\text{l}^2+1}{\text{l}}$
$\Rightarrow\ \sec\theta=\frac{\text{l}^2+1}{2\text{l}}$
Hence, proved.
$\sec\theta+\tan\theta=\text{l}\ \ [\text{Given}]\ (\text{I})$
$\Rightarrow\ (\sec\theta+\tan\theta)\frac{(\sec\theta-\tan\theta)}{\sec\theta-\tan\theta}=\text{l}$
$\Rightarrow\ \frac{\sec^2\theta-\tan^2\theta}{\sec\theta-\tan\theta}=\text{l}\ [\because1+\tan^2\theta=\sec^2\theta]$
$\Rightarrow\ \frac{1}{\sec\theta-\tan\theta}=\text{l}$
$\text{or}\ \sec\theta-\tan\theta=\frac{1}{\text{l}}\ \ (\text{II})$
Now, get $\sec\theta$ by eliminating $\tan\theta$ from (I) and (II).
It can be obtained by adding (I) and (II).
$\Rightarrow\ 2\sec\theta=\text{l}+\frac{1}{\text{l}}$
$\Rightarrow\ 2\sec\theta=\frac{\text{l}^2+1}{\text{l}}$
$\Rightarrow\ \sec\theta=\frac{\text{l}^2+1}{2\text{l}}$
Hence, proved.
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