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

The question is

" 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"

Explanation:

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

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

Which is the required answer.

Answer 2

Step-by-step explanation:

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