Question

If $\sec \theta -\tan \theta =x$, then show that $\sec \theta +\tan \theta =\frac{1}{x}$ and hence find the values of $\cos \theta$ and $\sin \theta$.

Answer 1

Given:

$\sec \theta -\tan \theta =x$ …………….$\left(1\right)$

To prove:

$\sec \theta +\tan \theta =1/x$

By trigonometric identity

$\Rightarrow {\sec }^2\theta -{\tan }^2\theta =1$

$\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$

Put value of $\left(1\right)$

$\Rightarrow x.(\sec \theta +\tan \theta )=1$

$\therefore \sec \theta +\tan \theta =1/x$

Now,

$\sec \theta -\tan \theta =x$ …………….$\left(1\right)$

$\sec \theta +\tan \theta =1/x$ ………….$\left(2\right)$

Add both equations

$\sec \theta -\tan \theta +\sec \theta +\tan \theta =x+\frac{1}{x}$

$\Rightarrow 2\cdot \sec \theta =\frac{x^2+1}{x}$

$\Rightarrow \sec \theta =\frac{x^2+1}{2x}$

$\Rightarrow \frac{1}{\cos \theta }=\frac{x^2+1}{2x}[\cos \theta =\frac{1}{\sec \theta }]$

$\Rightarrow \cos \theta =\frac{2x}{x^2+1}$

using trigonometric identify

$\Rightarrow \sin \theta =\sqrt{1-{\cos }^2\theta }$

$\Rightarrow \sin \theta =\sqrt{1-(2x/x^2+1{)}^2}$

$\Rightarrow \sin \theta =\sqrt{1-\frac{4x^2}{(x^2+1{)}^2}}$

$\Rightarrow \sin \theta =\sqrt{\frac{(x^2+1{)}^2-4x^2}{(x^2+1{)}^2}}$

$\Rightarrow \sin \theta =\sqrt{\frac{x^4+1-2x^2}{(x^2+1{)}^2}}$

$\Rightarrow \sin \theta =\sqrt{\frac{(x^2-1{)}^2}{(x^2+1{)}^2}}$

$\Rightarrow \sin \theta =\sqrt{{\left(\frac{x^2-1}{x^2+1}\right)}^2}$

$\therefore \sin \theta =\frac{x^2-1}{x^2+1}$

$\therefore \cos \theta =\frac{2x}{x^2+1}:\sin \theta =\frac{x^2-1}{x^2+1}$.