Question

If ${\displaystyle \sec \theta +\tan \theta =p,}$ prove that

(i)${\displaystyle \sec \theta =\frac{1}{2}(p+\frac{1}{p}) \left(ii\right)\tan \theta =\frac{1}{2}(p-\frac{1}{p})\left(iii\right)\sin \theta =\frac{p^2-1}{p^2+1}}$

Answer 1

(i) We have , ${\displaystyle \sec \theta +\tan \theta =p}$          ....................(1)

${\displaystyle \Rightarrow \frac{\sec \theta +\tan \theta }{1}\times \frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta }=p}$

${\displaystyle \Rightarrow \frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }=p}$

${\displaystyle \Rightarrow \frac{1}{\sec \theta -\tan \theta }=p}$

${\displaystyle \Rightarrow \sec \theta -\tan \theta =\frac{1}{p}}$       .........................(2)

Adding (1) and (2) , We get

2${\displaystyle \sec \theta =p+\frac{1}{p}}$

${\displaystyle \Rightarrow \sec \theta =\frac{1}{2}(p+\frac{1}{p})}$

(ii) subtracting (2) feom (1) , We get  

${\displaystyle 2\tan \theta =(p-\frac{1}{p})}$

${\displaystyle \Rightarrow \tan \theta =\frac{1}{2}(p-\frac{1}{p})}$

(iii) Using  (i) and (ii) , We get 

${\displaystyle \sin \theta =\frac{\tan \theta }{\sec \theta }}$

                =${\displaystyle \frac{\frac{1}{2}(p-\frac{1}{p})}{\frac{1}{2}(p+\frac{1}{p})}}$

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

∴ ${\displaystyle \sin \theta =\frac{p^2-1}{p^2+1}}$