Question

prove that

$\frac{ \sin( \alpha ) - \cos( \alpha ) + 1 }{ \sin( \alpha ) + \cos( \alpha ) - 1} = \frac{1}{ \sec( \alpha ) - \tan( \alpha ) }$

please solve this
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Answer 1

Solution

Taking LHS

$\sf\frac{ \sin( \alpha ) - \cos( \alpha ) + 1 }{ \sin( \alpha ) + \cos( \alpha ) - 1}$

= $\sf\frac{ \sin( \alpha ) - \cos( \alpha ) + 1 }{ \sin( \alpha ) + -(-\cos( \alpha ) + 1)}$

= $\sf\frac{ \sin( \alpha ) + 1 - \cos( \alpha )}{ \sin( \alpha ) -( 1 - \cos( \alpha ))}$

Multiply with $\sin( \alpha ) + 1 - \cos( \alpha )$ in both numerator and denominator

$\sf\frac{ (\sin( \alpha ) + 1 - \cos( \alpha ))(\sin( \alpha ) + 1 - \cos( \alpha ))}{ (\sin( \alpha ) -( 1 - \cos( \alpha ))(\sin( \alpha ) + 1 - \cos( \alpha ))}$

Using, (a + b)(a - b) = a² - b²

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

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

Using, (a + b - c)² = a² + b² + c² + 2ab - 2bc - 2ac

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

(since, sin²∅ + cos²∅ = 1 → sin²∅ - 1 = - cos²∅)

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

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

$\sf\frac{1(1 - \cos\alpha) + \sin\alpha(1 - \cos\alpha)}{\cos\alpha(1 - \cos\alpha)}$

$\sf\frac{(1- \cos\alpha)(\sin\alpha + 1)}{(\cos\alpha )(1 - \cos\alpha)}$

$\sf\frac{\sin\alpha + 1}{\cos\alpha}$

$\sf\frac{\sin\alpha}{\cos\alpha} + \frac{1}{\cos\alpha}$

$\sf \tan\alpha + \sec\alpha$

(since, sin∅/cos∅ = tan∅ and 1/cos∅ = sec∅)

Multiply with sec∅ - tan∅ in both numerator and denominator

$\sf\frac{(\sec\alpha + \tan\alpha)(\sec\alpha - \tan\alpha)}{(\sec\alpha - \tan\alpha}$

$\sf\frac{\sec^2\alpha - \tan^2\alpha}{\sec\alpha - \tan\alpha}$

Now, sec²∅ - tan²∅ = 1

$\implies\Large{\frac{1}{\sec\alpha - \tan\alpha}}$

Hence Proved.