Math, asked by namankhandelwal8999, 10 months ago



if sec^2(1+Sin)(1-Sin)=x ,then find the value of x.​

Answers

Answered by Anonymous
104

\huge\underline\mathfrak\blue{Answer-}

\huge\boxed{x\:=\:1}

_________________

\huge\underline\mathfrak\blue{Explanation-}

\leadsto { \sec(A) }^{2} (1 +  \sin(A) )(1 -  \sin(A) ) = x

By using,

★ (a + b)(a - b) = a² - b²

\leadsto \dfrac{1}{ { \cos(A) }^{2} } \times  ( {1}^{2}  -  { \sin(A) }^{2} ) = x

By using,

★ 1 - Sin²A = Cos²A

\leadsto \dfrac{1}{ { \cos(A) }^{2} }   \times  { \cos(A) }^{2} = x

By using,

\dfrac{1}{Cos(A)^{2}} = Sec²A

\leadsto \dfrac{1}{ {\cancel{\cos(A) }^{2} }}×{\cancel{\cos(A) }^{2}} = x

\leadsto 1 = x

\huge\leadsto \huge\boxed{x\:=\:1}

Answered by Anonymous
24

Solution :

 \sf \hookrightarrow { \sec }^{2}(a) \bigg(1 +  \sin(a) \bigg ) \bigg(1 -  \sin(a)  \bigg ) = x \\  \\  \sf \hookrightarrow   \frac{1}{ { \cos }^{2}(a)}   \bigg( {(1)}^{2}  -  { \sin}^{2} (a) \bigg) = x \\  \\  \sf \hookrightarrow  \frac{1}{ \cancel{{\cos }^{2}(a) }} \times { \cancel{{\cos }^{2}(a) }}  = x\\  \\  \sf \hookrightarrow  1 = x

Hence , the required value of x is 1

Identies used :

 \sf \star \:  \:  \sec(a) =  \frac{1}{ \cos(a) }   \\  \\  \sf \star \:  \:(a + b)(a - b) =   {a}^{2}  -  {b}^{2}  \\  \\  \sf \star \:  \: 1 -  { \sin}^{2} (a) =  { \cos}^{2} (a)

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