Question

$\sec \theta +\tan \theta =p$, then obtain the values of $\sec \theta ,\tan \theta$ and $\sin \theta$ in terms of p.

Answer 1

$\sec \theta +\tan \theta =p$

$\Rightarrow \frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=p$

$\Rightarrow 1+\sin \theta =p\cos \theta$

$\Rightarrow (1+\sin \theta {)}^2=p^2{\cos }^2\theta$

$\Rightarrow 1+{\sin }^2\theta +2\sin \theta =p^2-p^2{\sin }^2\theta$

$\Rightarrow 1+(1+p^2){\sin }^2\theta +2\sin \theta -p^2=0$

$\Rightarrow (1+p^2){\sin }^2\theta +2\sin \theta +1-p^2=0$

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

$=\frac{-2\pm \sqrt{4+p^4-1}}{2(1+p^2)}=\frac{-2\pm \sqrt{p^4+3}}{2(1+p^2)}$

$\Rightarrow \sec \theta +\tan \theta =p$

$\Rightarrow \frac{\sec \theta -\tan \theta =1/p}{2\sec \theta =p+\frac{1}{p}=\frac{1+p^2}{p}}$

$\Rightarrow \sec \theta =\frac{a+p^2}{2p}$

$\tan \theta =p-\sec \theta$

$=p-\left(\frac{1+p^2}{2p}\right)=\frac{p^2-1}{2p}$ .

Hence, this is the answer.