Question

prove that
coso.coto/1+sino = coseco -1​

Answer 1

Answer:-

We have to prove:-

$\dfrac{ \cos( \theta) \times \cot( \theta) }{1 + \sin( \theta) } = \csc( \theta) - 1$

using cot θ = cos θ/sin θ we get,

$\: \implies \:\frac{ \cos( \theta) \times \frac{ \cos( \theta) }{ \sin ( \theta)} }{1 + \sin( \theta) } = \csc( \theta) - 1 \\ \\ \\ \implies \: \frac{ \frac{ \cos ^{2} ( \theta) }{ \sin( \theta) } }{1 + \sin( \theta) } = \csc( \theta) - 1$

Using the identity cos² θ = 1 - sin² θ we get,

$\implies \: \dfrac{1 - { \sin}^{2} ( \theta)}{ \sin( \theta) } \times \dfrac{1}{1 + \sin( \theta) } = \csc( \theta) - 1$

Using a² - b² = (a + b)(a - b) in LHS we get,

$\implies \: \frac{(1 + \sin( \theta) )(1 - \sin( \theta)) }{ \sin( \theta) (1 + \sin( \theta)) } = \csc( \theta) - 1 \\ \\ \\ \implies \: \frac{1 - \sin( \theta) }{ \sin( \theta) } = \csc( \theta) - 1 \\ \\ \\ \implies \: \frac{1}{ \sin( \theta) } - \frac{ \sin( \theta) }{ \sin( \theta) } = \csc( \theta) - 1$

Using 1/sin θ = cosec θ in LHS we get,

$\implies \: \csc( \theta) - 1 = \csc( \theta) - 1$

Hence, Proved.