Question

If `sec theta + tan theta = p,` prove that

(i)`sec theta = 1/2 ( p+1/p)   (ii) tan theta = 1/2 ( p- 1/p) (iii) sin theta = (p^2 -1)/(p^2+1)`

Answer 1

(i) We have , `sec theta + tan theta = p`          ....................(1)

`⇒ (sec theta + tan theta )/1 xx (sec theta - tan theta )/( sec theta - tan theta ) = p`

`⇒ (sec ^2 theta - tan^2 theta )/( sec theta - tan theta) = p`

`⇒ 1/ (sec theta - tan theta ) =p`

`⇒ sec theta - tan theta = 1/ p`       .........................(2)

Adding (1) and (2) , We get

2` sec theta = p + 1/p`

`⇒ sec theta = 1/2 ( p+1/p)`

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

`2 tan theta = (p - 1/p)`

`⇒ tan theta = 1/2 ( p-1/p)`

(iii) Using  (i) and (ii) , We get 

`sin theta = tantheta/ sec theta`

                =`(1/2(p-1/p))/(1/2 (p+1/p)`

               =`(((p^2-1)/p))/(((p^2+1))/p)`

∴ `sin theta = (p^2-1)/(p^2 +1)`