Question

If $\sec \theta +\tan \theta =p$, then find the value $\text{cosec}\theta$.

Answer 1

$sec\theta +tan\theta =P$ _______ (1)

$se{c}^2\theta -ta{n}^2\theta =1$

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

$sec\theta -tan\theta =1/P$ _______ (2)

Subtracting (2) from (1), we get

$2tan\theta =P-\frac{1}{P}$

$\therefore cot\theta =\frac{2P}{P^2-1}$

Now, $cose{c}^2\theta -co{t}^2\theta =1$

$cose{c}^2\theta =1+co{t}^2\theta$

$\Rightarrow cose{c}^2\theta =1+(\frac{2P}{P^2-1}{)}^2$

$\Rightarrow cose{c}^2\theta =1+\frac{4P^2}{(P^2-1{)}^2}$

$\Rightarrow cose{c}^2\theta =(P^4-2P^2+1+4P^2)/(P^2-1{)}^2$

$\Rightarrow cose{c}^2\theta =\frac{P^2+1}{P^2-1}$