If #tantheta + sectheta = x#, show that #sintheta=(x^2-1)/(x^2+1)# ?

Question

1805 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-14T08:32:11

Given, #sectheta+tantheta=x....[1]#

#rarrsec^2theta-tan^2theta=1#

#rarr(sectheta+tantheta)(sectheta-tantheta)=1#

#rarrx*(sectheta-tantheta)=1#

#rarrsectheta-tantheta=1/x.....[2]#

Adding #[1] and [2]#,

#rarrsecthetacancel(+tantheta)+secthetacancel(-tantheta)=x+1/x#

#rarr2sectheta=(x^2+1)/x...[3]#

Subtracting #[2]# from #[1]#,we get

#rarrsectheta+tantheta-(sectheta-tantheta)=x-1/x#

#rarrcancel(sectheta)+tanthetacancel(-sectheta)+tantheta=x-1/x#

#rarr2tantheta=(x^2-1)/x....[4]#

Dividing #[4]# by #[3]#, we get,

#rarr(2tantheta)/(2sectheta)=((x^2-1)/x)/((x^2+1)/x)#

#rarrsintheta=(x^2-1)/(x^2+1)# Proved