Question

please solve the question correctly correctly.......DON'T SPAM......​

Answer 1

Answer:

your answer attached in the photo

Answer 2

Question :

To prove :

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

Formulas related to Trigonometry:

1. sin²A + cos²A = 1
2. sec²A - tan²A = 1
3. cosec²A - cot²A = 1
4. sin2A = 2 sinA cosA
5. cos2A = cos²A - sin²A
6. tan2A = 2 tanA / (1 - tan²A)

Solution :

LHS

$= \sqrt{ \frac{ \csc( \theta) - 1 }{ \csc( \theta) + 1 } }$

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

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

Now rationalise the denominator

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

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

we know that , sin²A + cos²A = 1

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

$= \frac{1 - \sin \theta }{ \cos \theta }$

RHS

$= \frac{1 - sin \theta}{cos \theta}$

⇒LHS = RHS

$\huge{\bold{ Hence \: Proved }}$