Question

Identity transformations of Trigonometric Expressions.
prove the following identities.
$\frac{sinx+cosx}{sinx-cosx}-\frac{se{c}^{2}x+2}{ta{n}^{2}x-1}=\frac{2}{tanx+1}.$

Answer 1

$LHS=\frac{sinx+cosx}{sinx-cosx}-\frac{se{c}^{2}x+2}{ta{n}^{2}x-1}$

= multiply & divide by

$(sinx+cosx)$

$=\frac{(sinx+cosx{)}^2}{-cos2x}-\left(\frac{2co{s}^{2}x+1}{si{n}^{2}x-co{s}^{2}x}\right)$

$=\frac{1+sin2x}{-cos2x}-\frac{2co{s}^{2}x+1}{(-cos2x)}$

$=\frac{1+sin2x-2co{s}^{2}x-1}{(-cos2x)}$

$=\frac{2sinxcosx-2co{s}^{2}x}{(-cos2x)}$

$=\frac{2cosx}{(-cos2x)}[sinx-cosx]$

$=\frac{2cosx}{si{n}^{2}x-co{s}^{2}x}[sinx-cosx]$

$=\frac{2cosx}{sinx+cosx}=\left(\frac{2}{tanx+1}\right)$

$=RHS$

Hence proved!