Math, asked by JamesonEmpire, 1 year ago

hey please do it by easy trick

both questions

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Answered by guptaramanand68
2
1:

 \frac{1 + \cos(x) + \sin(x) }{1 + \cos( x ) - \sin(x) } \\ \frac{1 + \cos(x) + \sin(x) }{1 + \cos( x ) - \sin(x)} \times \frac{1 + \cos(x) + \sin(x) }{1 + \cos(x) + \sin(x) } \\ \frac{ {(1 + \cos(x) + \sin(x) )}^{2} }{ {(1 + \cos( x )) }^{2} - \sin ^{2}x } \\ \frac{1 + { \cos }^{2}x + \sin ^{2} x + 2 \sin(x) \cos(x) + 2 \sin(x) + 2 \cos(x) }{1 + \cos ^{2} x + 2 \cos(x) - \sin ^{2}x } \\ \frac{2 + 2 \sin(x) + 2 \cos( x ) + 2 \sin(x) \cos(x) }{2 \cos ^{2}x + 2 \cos(x) } \\ \frac{1 + \sin(x) \cos(x) + \sin(x) + \cos(x) }{ \cos ^{2}x + \cos(x) } \\ \frac{1 + \cos(x) + \sin(x) (1 + \cos(x)) }{ \cos(x)(1 + \cos(x) ) } \\ \frac{(1 + \sin(x) )(1 + \cos(x)) }{ \cos(x)(1 + \cos(x)) } \\ \frac{1 + \sin(x) }{ \cos(x) }

2:

 \frac{ \cos(x) - \sin(x) + 1}{ \cos(x) + \sin(x) - 1 } \\

Divide both numerator and denominator by sin(x)

 \frac{ \cot(x) - 1 + \csc(x) }{ \cot(x) + 1 - \csc(x) } \\ \frac{ \cot(x) + \csc(x) - 1 }{ \cot(x) - ( \csc(x) - 1) } \\ \frac{ \cot(x) + \csc(x) - 1 }{ \cot(x) - ( \csc(x) - 1) } \times \frac{ \cot(x) + \csc(x) - 1 }{ \cot(x) + \csc(x) - 1 } \\ \frac{( \cot(x) + \csc(x) - 1) ^{2} }{ { \cot }^{2} x - {( \csc(x) - 1)}^{2} } \\ \frac{ { \cot }^{2} x + { \csc }^{2} x + 1 + 2 \cot(x) \csc(x) - 2 \csc(x) - 2 \cot(x) }{ \cot^{2} x - { \csc }^{2} x - 1 + 2 \csc(x) } \\ \frac{2 \csc^{2} x + 2 \cot(x) \csc(x) - 2 \csc(x) - 2 \cot(x) }{2 \csc(x) - 2} \\ \frac{ \csc ^{2} x + \cot(x) \csc(x) - \csc(x) - \cot(x) }{ \csc(x) - 1 } \\ \frac{ \csc(x) ( \csc(x) + \cot(x) ) - ( \csc(x) + \cot(x)) }{ \csc(x) - 1 } \\ \frac{( \csc(x) - 1)( \csc(x) + \cot(x)) }{( \csc(x) - 1) } \\ \csc(x) + \cot(x)


Done.

JamesonEmpire: hey why u divide by sin x
guptaramanand68: Dividing by sin(x) would bring cosec(x) and cot(x) into the expression straight away. There is no hard and fast rule to divide by sin(x), but I like this way.
guptaramanand68: If you want to do without dividing by sin(x), you can do it, there won't be any problem
JamesonEmpire: okk thanks bro
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