Question

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

Answer 1

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

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

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

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

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

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

$\leadsto$ $1$ = $x$

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

Answer 2

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 :

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