Question

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From Trigonometry class 10 ...

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$\frac{ \cos( \alpha ) - \sin( \alpha ) + 1 }{ \cos( \alpha ) + \sin( \alpha ) - 1 } = cosec ( \alpha ) + \cot( \alpha ) \\ \\ using \: the \: identity \: {cosec}^{2} ( \alpha) = 1 + \cot( \alpha )$


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Answer 1

Hey Miss Perfect ÷_÷

(•) Trigonometry ✓✓

Given Expression,

$E = \frac{ \cos( \alpha ) - \sin( \alpha ) + 1 }{ \cos( \alpha ) + \sin( \alpha ) - 1 }$

* But the Formula,

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

-> ReQuires, the basic terms cosec and cot !

So, Divide and Multiply the expression by sin

$E = \frac{ \cot( \alpha ) + \csc( \alpha ) - 1 }{ \cot( \alpha ) - \csc( \alpha ) + 1 }$

Now, substitute the value of "1" ( from the formula ) in Numerator !

$E = \frac{( \cot( \alpha ) + \csc( \alpha ))(1 + \cot( \alpha ) - \csc( \alpha ) )}{(1 + \cot( \alpha ) - \csc( \alpha ) )}$

Cancel the common terms and ->

$E = ( \cot( \alpha ) + \csc( \alpha ))$

Answer 2

heya....!!!

✔here is ua answer:

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E=cos(α)+sin(α)−1cos(α)−sin(α)+1​ 

* But the Formula,

\csc^{2} ( \alpha ) = 1 + \cot {}^{2} ( \alpha )csc2(α)=1+cot2(α) 

-> ReQuires, the basic terms cosec and cot ! 

So, Divide and Multiply the expression by sin

E = \frac{ \cot( \alpha ) + \csc( \alpha ) - 1 }{ \cot( \alpha ) - \csc( \alpha ) + 1 }E=cot(α)−csc(α)+1cot(α)+csc(α)−1​ 

Now, substitute the value of "1" ( from the formula ) in Numerator !

E = \frac{( \cot( \alpha ) + \csc( \alpha ))(1 + \cot( \alpha ) - \csc( \alpha ) )}{(1 + \cot( \alpha ) - \csc( \alpha ) )}E=(1+cot(α)−csc(α))(cot(α)+csc(α))(1+cot(α)−csc(α))​ 

Cancel the common terms and ->

E = ( \cot( \alpha ) + \csc( \alpha ))E=(cot(α)+csc(α))

hope it helps...!!!!❤