Question

prove that :

please solve it with explanation​

Answer 1

Answer:-

We have to prove:

$\sf \: \big( \sec(A) - \tan(A) \big) ^{2} = \dfrac{1 - \sin(A) }{1 + \sin(A) }$

Using sec A = 1/cos A and tan A = sin A/cos A in LHS we get;

$\implies \sf \: \bigg( \frac{1}{ \cos(A) } - \frac{ \sin(A) }{ \cos(A) } \bigg) ^{2} = \frac{1 - \sin(A) }{1 + \sin(A) } \\ \\ \\ \implies \sf \:\bigg( \frac{1 - \sin(A) }{ \cos(A) } \bigg) ^{2} =\frac{1 - \sin(A) }{1 + \sin(A) } \\ \\ \\ \implies \sf \:\frac{ \big( {1 - \sin(A) \big)}^{2} }{ { \cos }^{2}(A) } = \frac{1 - \sin(A) }{1 + \sin(A) }$

We know that;

sin² A + cos² A = 1

⟹ cos² A = 1 - sin² A

using a² - b² = (a + b)(a - b) in RHS we get,

⟹ cos² A = (1 + sin A)(1 - sin A)

Substituting the value of cos² A in LHS we get;

$\implies \sf \: \frac{(1 - \sin(A) ) \cancel{(1 - \sin(A))} }{(1 + \sin(A)) \cancel{(1 - \sin(A) )}} = \frac{1 - \sin(A) }{1 + \sin(A) } \\ \\ \\ \implies \sf \:\frac{1 - \sin(A) } {1 + \sin(A) } = \frac{1 - \sin(A) }{1 + \sin(A) }$

Hence, Proved.