Question

√1+sin a/√1-sin a=sec a+tan au​

Answer 1

Answer:

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Answer 2

Question :

To prove

$\sqrt{ \frac{1 + \sin(a) }{1 - \sin(a) } } = \sec(a) + \tan(a)$

Trignometric Formulas:

1. sin²A + cos²A = 1
2. sec²A - tan²A = 1
3. cosec²A - cot²A = 1
4. sec a =$\frac{1}{cosa}$
5. cosec a=$\frac{1}{sina}$
6. cot a = $\frac{1}{tana}$

Solution :

LHS

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

Rationalise the denominator

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

we know that a²- b² = (a+b)(a-b

___________

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

we know ;sin²A +cos²A = 1

$= \sqrt{ \frac{(1 + \sin(a)) {}^{2} }{ \cos {}^{2} (a) } }$

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

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

$= \sec(a) + \tan(a)$

RHS = sec a + tan a

⇒ LHS = RHS

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

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More Trigonometry Formulas

1. sin2A = 2 sinA cosA
2. cos2A = cos²A - sin²A
3. tan2A = 2 tanA / (1 - tan²A)