Question

prove that sin theta minus cos theta upon sin theta + cos theta + sin theta + cos theta upon sin theta minus cos theta is equal to 2 upon sin squared theta minus one

Answer 1

Answer:this is coming

Step-by-step explanation:

Answer 2

The correct question is

Prove that

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

(Note : Representing theta as angle $\alpha$ )

Solution:

We have to prove that

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

• We have, the trigonometric equation:

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

• Solving LHS :

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

• Rationalizing both the terms that is multiplying and dividing both the terms by (sin$\alpha$ - cos$\alpha$ ) and (sin$\alpha$ + cos$\alpha$) respectively, we get

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

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

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

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

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

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

• Using the identity

$sin^2\alpha +cos^2\alpha = 1$, we get

$=\frac{2 }{sin^{2}\alpha - cos^{2} \alpha }$

• Using the identity

$cos^2\alpha = 1- sin^2\alpha$, we get

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

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

= RHS

  therefore

$\frac{sin\alpha- cos\alpha }{sin\alpha + cos\alpha } + \frac{sin\alpha + cos\alpha }{sin\alpha - cos\alpha } = \frac{2}{2sin^{2}\alpha -1 }$, hence proved.