If \(\tan\theta+\sec\theta=\ell\), prove that \(\sec\theta=\dfrac{\ell^2+1}{2\ell}\).
\(\sec\theta=\dfrac{\ell^2+1}{2\ell}\).
Step 1: We are given that \(\tan\theta + \sec\theta = \ell\).
Step 2: Recall the identity: \((\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1\). This means that if we know \(\sec\theta + \tan\theta\), then we can also find \(\sec\theta - \tan\theta\).
Step 3: Since \(\sec\theta + \tan\theta = \ell\), substitute this into the identity:
\(\ell(\sec\theta - \tan\theta) = 1\)
So, \(\sec\theta - \tan\theta = \dfrac{1}{\ell}\).
Step 4: Now we have two equations:
Step 5: Add these two equations to remove \(\tan\theta\):
\((\sec\theta + \tan\theta) + (\sec\theta - \tan\theta) = \ell + \dfrac{1}{\ell}\)
\(2\sec\theta = \ell + \dfrac{1}{\ell}\)
Step 6: Divide both sides by 2 to isolate \(\sec\theta\):
\(\sec\theta = \dfrac{\ell + \dfrac{1}{\ell}}{2}\)
Step 7: Simplify the fraction:
\(\sec\theta = \dfrac{\dfrac{\ell^2 + 1}{\ell}}{2}\)
\(\sec\theta = \dfrac{\ell^2 + 1}{2\ell}\)
Final Answer: \(\sec\theta = \dfrac{\ell^2 + 1}{2\ell}\).