Question

prove that :1 - sin2A/(1+cosA) = cos A.....no spamming...✖️☠️​

Answer 1

Required Answer:-

Given to prove:

$\rm \mapsto 1 - \dfrac{ { \sin}^{2}(x) }{1 + \cos(x) } = \cos(x)$

Proof:

Taking LHS,

$\rm 1 - \dfrac{ { \sin}^{2}(x) }{1 + \cos(x) }$

$\rm = \dfrac{1 + \cos(x) - { \sin}^{2}(x) }{1 + \cos(x) }$

Now, we know that,

➡ sin²(x) + cos²(x) = 1

➡ sin²(x) = 1 - cos²(x)

$\rm = \dfrac{1 + \cos(x) - (1 - \cos^{2} (x) ) }{1 + \cos(x) }$

Applying formula a² - b² = (a + b)(a - b), we get,

$\rm = \dfrac{1 + \cos(x) - (1 + \cos(x))(1 - \cos(x)) }{1 + \cos(x) }$

Now, taking 1 + cos(x) as common, we get,

$\rm = \dfrac{(1 + \cos(x)) \{1 - (1 - \cos(x) \}}{1 + \cos(x) }$

$\rm = \dfrac{(1 + \cos(x)) \{ \cos(x) \}}{1 + \cos(x) }$

$\rm = \cos(x)$

= RHS (Hence Proved)

Answer 2

Answer:

$\huge\boxed{\fcolorbox{green} {black}{\color{yellow}{REQUIRED ANSWER —}}}$

$\boxed{\boxed{1 - \frac{ { \sin( a ) }^{2} }{1 + \cos(a) } = \cos(a) }}$

$\huge\boxed{\fcolorbox{purple} {black}{\color{yellow}{PROOF — }}}$

Taking LHS :

$</u></strong><strong><u>\</u></strong><strong><u>large</u></strong><strong><u> 1 - \frac{ { \sin( a ) }^{2} }{1 + \cos(a) }$

Taking LCM

$</strong><strong>\</strong><strong>large</strong><strong> = \frac{1 + \cos(a) - { \sin(a) }^{2} }{1 + \cos(a) }$

$</strong><strong>\</strong><strong>large</strong><strong> = \frac{1 + \cos(a) - (1 - { \cos(a) )}^{2} }{1 + \cos(a) }$

$Applying \: \: a²-b² \: = \: (a+b)(a-b)$

$</strong><strong>\</strong><strong>large</strong><strong> = \frac{1 + \cos(a) - (1 + \cos(a) )(1 - \cos(a)) }{1 + \cos(a) }$

$Now \: \: we \: \: take \: \: 1+cos(a) \: \: common</strong></p><p><strong>$

$</strong><strong>\</strong><strong>large</strong><strong> = \frac{(1 + \cos(a) )(1 - (1 - \cos(a) ))}{1 + \cos(a) }$

1+ cos(a) cancels out

$</strong><strong>\</strong><strong>large</strong><strong> = (1 - 1 + \cos(a) ) = \cos(a)$

.....LHS = RHS

....................(Hence Proved)

_______________

Trignometrical Ratios Identites(one is used above as well)are :–

${\implies \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1$

$\implies1 + {\tan( \alpha ) }^{2} = { \sec( \alpha ) }^{2}$

$\implies 1 + { \cot( \alpha ) }^{2} = { \csc( \alpha ) }^{2}$

___________________

Mark as BRAINLIEST

${\huge{\underline{\overline{\rm{\color{yellow}{\longrightarrow @phenom \longleftarrow}}}}}}</p><p></p><p>$