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

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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

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