Question

Prove that : cos A / (1-tanA) + sinA /( 1-cotA) = sinA + cos A ​

Answer 1

★ Given :-

• $\sf \: \dfrac{ \cos(A) }{1 - \tan(A) } + \dfrac{ \sin(A) }{1 - \cot(A) }$

★ To Prove :-

• $\sf \: \dfrac{ \cos(A) }{1 - \tan(A) } + \dfrac{ \sin(A) }{1 - \cot(A) } = \sin(A) + \cos(A)$

★ Solution :-

$\sf \:LHS = \dfrac{ \cos(A) }{1 - \tan(A) } + \dfrac{ \sin(A) }{1 - \cot(A) }$

$= \sf \: \dfrac{ \cos(A) }{1 - \dfrac{ \sin(A) }{ \cos(A) } } + \dfrac{ \sin(A) }{1 - \dfrac{ \cos(A) }{ \sin(A) } }$

$= \sf \: \dfrac{ \cos(A) }{ \dfrac{ \cos(A) - \sin(A) }{ \cos(A) } } + \dfrac{ \sin(A) }{ \dfrac{ \sin(A) - \cos(A) }{ \sin(A) } }$

$= \sf \: \dfrac{ { \cos(A) }^{2} }{ \cos(A) - \sin(A) } + \dfrac{ { \sin(A) }^{2} }{ \sin(A) - \cos(A) }$

$= \sf \: \dfrac{ { \cos(A) ^{2} - \sin(A) }^{2} }{ \cos(A) - \sin(A) }$

[by using a^2 - b^2 = (a + b) (a - b)]

$= \sf \: \dfrac{[ \cos(A) + \sin(A) ] [ \cos(A) - \sin(A) ]}{ \cos(A) - \sin(A) }$

$= \sf \: \cos(A) + \sin(A) = RHS (Proved)$

Answer 2

$\sf\dfrac{CosA}{1-tanA}$ + $\sf\dfrac{sinA}{1-CotA}$

$\sf\dfrac{CosA}{1-\sf\dfrac{SinA}{CosA}}$ + $\sf\dfrac{SinA}{1-\sf\dfrac{CosA}{SinA}}$

$\sf\dfrac{CosA}{\sf\dfrac{CosA-SinA}{CosA}}$ + $\sf\dfrac{SinA}{\sf\dfrac{SinA-CosA}{SinA}}$

$\sf\dfrac{Cos²A}{CosA-SinA}$ + $\sf\dfrac{sin²A}{SinA-CosA}$

$\sf\dfrac{Cos²A}{CosA-SinA}$ - $\sf\dfrac{sin²A}{CosA-SinA}$

$\sf\cancel\dfrac{(CosA+SinA)(CosA-SinA)}{CosA-SinA}$

$\sf{\ CosA + SinA }$ = R.H.S