Question

Prove that sin theta + cos theta upon sin theta minus cos theta + sin theta minus cos theta sin theta + cos theta equals to 2 sin squared theta minus cos squared theta equals to 2 upon 2 sin squared theta minus one

Answer 1

Answer:

Step-by-step explanation:

$\frac{sin\theta+cos\theta}{sin\theta-cos\theta}+\frac{sin\theta-cos\theta}{sin\theta+cos\theta}$

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

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

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

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

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

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

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

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

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

Answer 2

Answer:

Step-by-step explanation:

The given equation is :

$\frac{sin\theta+cos\theta}{sin\theta-cos\theta}+\frac{sin\theta-cos\theta}{sin\theta+cos\theta}$

Solving this equation, we get

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

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

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

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

Hence proved.