Question

Prove that (cosA - sinA + 1) / (cosA + sinA - 1) = (cosA +1) / sinA​

Answer 1

Solution:-

We have

$\rm \implies \dfrac{ \cos A- \sin A + 1}{ \cos A+ \sin A - 1} = \dfrac{ \cos A + 1}{ \sin A }$

Using cross multiplication methods

$\rm \implies \: \sin A( \cos A - \sin A + 1) = (\cos A + 1)( \cos A + \sin A - 1)$

$\rm \implies \: \sin A \cos A - \sin {}^{2} A + \sin A = \cos {}^{2} A + \sin A \cos A - \cos A + \cos A + \sin A - 1$

$\rm \implies \: \cancel{ \sin A \cos A} - \sin {}^{2} A + \cancel{\sin A} = \cos {}^{2} A + \cancel{\sin A \cos A} - \cancel{\cos A }+ \cancel{ \cos A} + \cancel{\sin A} - 1$

$\rm \implies \: - \sin {}^{2} A = \cos {}^{2} A - 1$

$\rm \implies \: \cos {}^{2} A + \sin {}^{2} A = 1$

We know that

$\rm \implies \: \cos {}^{2} A + \sin {}^{2} A = 1$

We get

$\rm \implies \: 1 = 1$

Hence proved

Some more trigonometry identities

$\rm \implies \: \cos {}^{2} A + \sin {}^{2} A = 1$

$\rm \implies \: \tan \theta = \dfrac{ \sin \theta }{ \cos\theta}$

$\rm \implies \cot \theta = \dfrac{ \cos \theta }{ \sin \theta }$

$\rm \implies \: \sec \theta = \dfrac{1}{ \cos \theta }$

$\implies \csc \theta = \dfrac{1}{ \sin\theta}$