Question

sin𝜃+cos𝜃= 1−2cos²𝜃/sin 𝜃−cos 𝜃

Answer 1

$sin\theta+cos\theta=\frac{1-2cos^2\theta}{sin\theta-cos\theta}$

Rearranging the right hand side(RHS) becomes,

$\frac{-1(2cos^2\theta-1)}{-1(cos \theta-sin\theta)}=\frac{2cos^2\theta-1}{cos\theta-sin\theta}$

From the following trigonometric identity,

$cos^2\theta+sin^2\theta=1$

then, the numerator,

$2cos^2\theta-1=2cos^2\theta-(cos^2\theta+sin^2\theta)$

so,

$2cos^2\theta-1=cos^2\theta-sin^2\theta$

Substituting for the numerator in the RHS,

$sin\theta+cos\theta=\frac{cos^2\theta-sin^2\theta}{cos\theta-sin\theta}$

The numerator of the right hand side is a difference of two squares, thus,

$cos^2\theta-sin^2\theta=(cos\theta+sin\theta)(cos\theta-sin\theta)$

Substituting for the numerator of the RHS,

$sin\theta+cos\theta=\frac{(cos\theta+sin\theta)(cos\theta-sin\theta)}{cos\theta-sin\theta}$

Simplifying the RHS proves that the left hand side is equal to the right hand side, that is,

$sin\theta+cos\theta=cos\theta+sin\theta$

This proves that,

$sin\theta+cos\theta=\frac{1-2cos^2\theta}{sin\theta-cos\theta}$