Question

$If\sec x+\tan x=k,thenprovethat\sin x=\frac{k^2-1}{k^2+1}.$

Answer 1

We have $secx+tanx=k...........1$

From 2nd identity

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

$a^2-b^2=(a+b)(a-b)$

$(secx+tanx)(secx-tanx)=1$

$\left(k\right)(secx-tanx)=1$ from eq 1

$secx-tanx=\frac{1}{k}............2$

adding equations 1 and 2, then we get

$secx+tanx+secx-tanx=1+\frac{1}{k}$

$secx+secx=1+\frac{1}{k}$

$2secx=k^2+\frac{1}{k}$

$secx=k^2+\frac{1}{2k}$

we know that $secx=\frac{1}{cosx}$

$cosx={\frac{1}{k}}^2+\frac{1}{2k}$

$cosx=\frac{2k}{k^2+1}$

we know that from 1st identity

$sinx=\sqrt{(1-cos{x}^2)}$

$sinx=\sqrt{(1-(\frac{2k}{(k^2+1{)}^2}})$

$sinx=\sqrt{\frac{(1-4k^2)}{(k^2+1{)}^2}}$

$sinx=\sqrt{\frac{(k^2+1{)}^2-4k^2}{k^2+1}}$

$sinx=\sqrt{\frac{(k^2-1{)}^2}{(k^2+1{)}^2}}$

$sinx=\frac{k^2-1}{k^2+1}$

Hence proved