Question:
If $\tan x=t$ then $\tan 2 x+\sec 2 x=$
(a) $\frac{1+t}{1-t}$
(b) $\frac{1-t}{1+t}$
(c) $\frac{2 t}{1-t}$
(d) $\frac{2 t}{1+t}$
Solution:
(a) $\frac{1+t}{1-t}$
$\tan 2 x+\sec 2 x=\frac{2 \tan x}{1-\tan ^{2} x}+\frac{1+\tan ^{2} x}{1-\tan ^{2} x}$
$=\frac{2 \tan x+1+\tan ^{2} x}{1-\tan ^{2} x}$
$=\frac{(1+\tan x)^{2}}{1-\tan ^{2} x}$
$=\frac{(1+\tan x)(1+\tan x)}{(1+\tan x)(1-\tan x)}$
$=\frac{1+\tan x}{1-\tan x}$
$=\frac{1+t}{1-t} \quad[\tan x=t$ (given) $]$
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