MCQ
If $\sec\theta+\tan\theta=\text{x}\sec\theta+\tan\theta=\text{x},$ then $\tan\theta=\tan\theta=$
- A$\frac{\text{x}^2+1}{\text{x}}$
- B$\frac{\text{x}^2-1}{\text{x}}$
- C$\frac{\text{x}^2+1}{2\text{x}}$
- ✓$\frac{\text{x}^2-1}{2\text{x}}$
✓
Answer
Correct option: D.
$\frac{\text{x}^2-1}{2\text{x}}$
$\sec\theta+\tan\theta=\text{x}\ .....\text{(i)}$
We know that
$\sec^2\theta-\tan^2\theta=1$
$\Rightarrow \ (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$
$\Rightarrow\ \text{x}(\sec\theta-\tan\theta)=1$
$\Rightarrow\ \sec\theta-\tan\theta=\frac{1}{\text{x}}\ .....(\text{ii})$
Subtracting (ii) from (i)
$2\tan\theta=\text{x}-\frac{1}{\text{x}}=\frac{\text{x}^2-1}{\text{x}}$
$\tan\theta=\frac{\text{x}^2-1}{2\text{x}}$
We know that
$\sec^2\theta-\tan^2\theta=1$
$\Rightarrow \ (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1$
$\Rightarrow\ \text{x}(\sec\theta-\tan\theta)=1$
$\Rightarrow\ \sec\theta-\tan\theta=\frac{1}{\text{x}}\ .....(\text{ii})$
Subtracting (ii) from (i)
$2\tan\theta=\text{x}-\frac{1}{\text{x}}=\frac{\text{x}^2-1}{\text{x}}$
$\tan\theta=\frac{\text{x}^2-1}{2\text{x}}$
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