If \( \tan x=t \), Find \( \tan2x+\sec2x \)
Question
If
\[ \tan x=t, \]
then
\[ \tan2x+\sec2x \]
is equal to
(a) \(\dfrac{1+t}{1-t}\)
(b) \(\dfrac{1-t}{1+t}\)
(c) \(\dfrac{2t}{1-t}\)
(d) \(\dfrac{2t}{1+t}\)
Solution
Using the identities
\[ \tan2x=\frac{2t}{1-t^2} \]
and
\[ \sec2x=\frac{1+t^2}{1-t^2} \]
Therefore,
\[ \tan2x+\sec2x = \frac{2t}{1-t^2} + \frac{1+t^2}{1-t^2} \]
\[ = \frac{1+2t+t^2}{1-t^2} \]
\[ = \frac{(1+t)^2}{(1-t)(1+t)} \]
\[ = \frac{1+t}{1-t} \]
Final Answer
\[ \boxed{\frac{1+t}{1-t}} \]
Hence, the correct option is (a) \(\dfrac{1+t}{1-t}\).