Question

Prove the following with the help of identities:

${}{ \bf \frac{ \sin(theta) + 1 - \cos(theta) }{ cos(theta) - 1 + \sin(theta) } = \frac{1 + \sin(theta) }{ \cos(theta) } }$
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Answer 1

AnswEr-

Consider LHS =

$\frac{ \sin(theta) + 1 - \cos(theta) }{ \cos(theta) - 1 + \sin(theta) }$

Multiply and divide by {cos (theta)+1+sin(theta)}

$\frac{ \sin(theta ) + 1 - \cos(theta) }{ \cos(theta) - 1 + \sin(theta) } \times \frac{ \cos(theta) + 1 + \sin(theta) }{ \cos(theta) + 1 + \sin(theta) }$

$= \frac{(1 + sin(theta)){}^{2} - {cos}^{2} (theta)}{ (\ \sin(theta) + cos(theta)) {}^{2} - 1}$

$= \frac{1 + { \sin }^{2} (theta) + 2 \sin(theta) {} - 1(1 - \sin {}^{2} (theta) }{ { \sin }^{2}theta + \cos {}^{2} (theta) + 2 \cos(theta) sin(theta) - 1 }$

$= \frac{2 \sin(theta) (1 + \sin(theta) }{2 \sin. \cos }$

${}{{ = \frac{1 + \sin(theta) }{cos(theta)} }}$

=RHS .

Thus proved.