Question

1/(Cos A + Sin A -1)+1/( CosA + sin A + 1 )=cosec A + sec A​

Answer 1

Given:

$\dfrac{1}{\cos \:A + \sin A - 1} + \dfrac{1}{\cos \:A + \sin A + 1} = \cosec \:A + \sec \: A$

Solution:

Taking LHS:

$\dfrac{1}{\cos \:A + \sin A - 1} + \dfrac{1}{\cos \:A + \sin A + 1}$

Now taking LCM,

$\large{⇒}$$\dfrac{\cos \:A + \sin A +1 +\cos A \: + \sin \: A - 1}{(\cos \:A + \sin A - 1) (\cos \:A + \sin A + 1) }$

$\large{⇒}$$\dfrac{2\cos \:A +2\sin \:A }{((\cos A \: + \sin \: A) - 1)((\cos A \: + \sin \: A) + 1)}$

$\large{⇒}$$\dfrac{2(\cos \:A +\sin \:A ) }{((\cos A \: + \sin \: A)^{2} - 1^{2} )}$

$\large{⇒}$$\dfrac{2(\cos \:A +\sin \:A ) }{(\cos^{2} A \: + \sin ^{2} \: A + 2\sin A\: \cos \: A- 1)}$

$\large{⇒}$$\dfrac{2(\cos \:A +\sin \:A ) }{1 + 2\sin A\: \cos \: A- 1)}$

$\large{⇒}$$\dfrac{2(\cos \:A +\sin \:A ) }{2(\sin A\: \cos \: A)}$

$\large{⇒}$$\dfrac{(\cos \:A +\sin \:A ) }{( \sin A\: \cos \: A)}$

$\large{⇒}$$\dfrac{\cos \:A}{(\sin A\: \cos \: A)}+\dfrac{\sin \:A}{(\sin A\: \cos \: A)}$

$\large{⇒}$$\dfrac{1}{\sin A} + \dfrac{1}{ \cos \: A}$

$\large{⇒}$$\cosec A\: + \sec \: A$

LHS = RHS

LHS = RHSProved

Formulas Used:

$\sf {\sin}^2 A + {\cos}^2 A = 1$

$\sf {(a+b)}^2 = a^2 + b^2 + 2ab$