Question

If x=1+sin a/cos a then sin a equal
a.1-x^2/1+x^2
b.x^2-1/x^2+1
C.1+X/1-x
D.1-x/1+x

Answer 1

Step-by-step explanation:

We have,

$\rm \: x = \frac{1 + \sin( \alpha ) }{ \cos( \alpha ) }$

$\rm \: \implies x \cos( \alpha ) = 1 + \sin( \alpha )$

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

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

$\rm \: \implies x^{2}(1 - \sin ^{2} ( \alpha ) )= 1 + \sin^{2} ( \alpha ) + 2 \sin( \alpha )$

$\rm \: \implies x^{2}- {x}^{2} \sin ^{2} ( \alpha ) = 1 + \sin^{2} ( \alpha ) + 2 \sin( \alpha )$

$\rm \: \implies - x^{2} + {x}^{2} \sin ^{2} ( \alpha ) + 1 + \sin^{2} ( \alpha ) + 2 \sin( \alpha ) = 0$

$\rm \: \implies ( {x}^{2} + 1) \sin ^{2} ( \alpha ) + 2 \sin( \alpha ) + (1 - {x}^{2}) = 0$

$\implies \sin( \alpha ) = \frac{ - 2 \pm \sqrt{( - 2)^{2} + 4( {x}^{2} + 1)( {x}^{2} - 1)} }{2( {x}^{2} + 1) }$

$\implies \sin( \alpha ) = \frac{ - 2 \pm \sqrt{4 + 4( {x}^{4} - 1)} }{2( {x}^{2} + 1) }$

$\implies \sin( \alpha ) = \frac{ - 2 \pm \sqrt{4 + 4{x}^{4} - 4} }{2( {x}^{2} + 1) }$

$\implies \sin( \alpha ) = \frac{ - 2 \pm \sqrt{ 4{x}^{4} } }{2( {x}^{2} + 1) }$

$\implies \sin( \alpha ) = \frac{ - 2 \pm 2{x}^{ 2} }{2( {x}^{2} + 1) }$

$\implies \sin( \alpha ) = \frac{ - 2 + 2{x}^{ 2} }{2( {x}^{2} + 1) }\: \: or \: \: \sin( \alpha ) = \frac{ - 2 - 2{x}^{ 2} }{2( {x}^{2} + 1) }$

$\implies \sin( \alpha ) = \frac{2( {x}^{ 2} - 1) }{2( {x}^{2} + 1) }\: \: or \: \: \sin( \alpha ) = \frac{ - 2 (1 + {x}^{ 2} ) }{2( {x}^{2} + 1) }$

$\implies \sin( \alpha ) = \frac{( {x}^{ 2} - 1) }{( {x}^{2} + 1) }\: \: or \: \: \sin( \alpha ) = \frac{ - 2 }{2}$

$\implies \sin( \alpha ) = \frac{( {x}^{ 2} - 1) }{( {x}^{2} + 1) }\: \: or \: \: \sin( \alpha ) = - 1$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:x = \dfrac{1 + sina}{cosa}$

On squaring both sides, we get

$\rm :\longmapsto\: {x}^{2} = \dfrac{ {(1 + sina)}^{2} }{ {cos}^{2} a}$

can be rewritten as

$\rm :\longmapsto\:\dfrac{ {x}^{2} }{1} = \dfrac{ {(1 + sina)}^{2} }{ {cos}^{2} a}$

On applying Componendo and Dividendo, we get

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

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + {sin}^{2}a + 2sina + {cos}^{2} a }{ 1 + {sin}^{2}a + 2sina - {cos}^{2} a}$

can be re-arranged as

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

We know,

$\boxed{ \rm{ {sin}^{2}x + {cos}^{2}x = 1}}$

So, using this

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1 + 1+ 2sina }{ {sin}^{2}a + {sin}^{2}a + 2sina}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{2+ 2sina }{2{sin}^{2}a + 2sina}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{2(1+ sina)}{2{sina}(sina + 1)}$

$\rm :\longmapsto\:\dfrac{ {x}^{2} + 1}{ {x}^{2} - 1} = \dfrac{1}{sina}$

$\bf\implies \:sina = \dfrac{ {x}^{2} - 1}{ {x}^{2} + 1}$

Hence, Option (b) is correct

Additional Information :-

$\red{\rm :\longmapsto\:If \: \dfrac{a}{b} = \dfrac{c}{d} , \: then}$

$\boxed{ \rm{ \frac{a}{c} = \frac{b}{d} \: is \: called \: alternendo}}$

$\boxed{ \rm{ \frac{b}{a} = \frac{d}{c} \: is \: called \: invertendo}}$

$\boxed{ \rm{ \frac{a + b}{b} = \frac{c + d}{d} \: is \: called \: componendo}}$

$\boxed{ \rm{ \frac{a - b}{b} = \frac{c - d}{d} \: is \: called \: dividendo}}$

$\boxed{ \rm{ \frac{a + b}{a - b} = \frac{c + d}{c - d} \: is \: called \: componendo \: and \: dividendo}}$