MCQ
If $\tan\text{x}=\text{t}$ then $\tan2\text{x}+\sec2\text{x}$ is equal to:
- A$\frac{1+\text{t}}{1-\text{t}}$
- B$\frac{1-\text{t}}{1+\text{t}}$
- C$\frac{2\text{t}}{1-\text{t}}$
- D$\frac{2\text{t}}{1+\text{t}}$
Solution:
$\tan2\text{x}+\sec2\text{x}=\frac{2\tan\text{x}}{1-\tan^2\text{x}}+\frac{1+\tan^2\text{x}}{1-\tan^2\text{x}}$
$=\frac{2\tan\text{x}+1\tan^2\text{x}}{1-\tan^2\text{x}}$
$=\frac{(1+\tan\text{x})^2}{1-\tan^2\text{x}}$
$=\frac{(1+\tan\text{x})(1+\tan\text{x})}{(1+\tan\text{x})(1-\tan\text{x})}$
$=\frac{1+\tan\text{x}}{1-\tan\text{x}}$
$\frac{1+\text{t}}{1-\text{t}}$ $[\tan\text{x}=\text{t}]$
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