Question

1/1 - sinA + 1/1 + sinA = 2 sec^2

Answer 1

Answer :-

We have :

$\rightarrow \bold{\dfrac{1}{1-sinA}+\dfrac{1}{1+sinA} }$

Taking L.C.M :

$\rightarrow \bold{\dfrac{(1+sinA)+(1-sinA)}{(1-sinA)(1+sinA)} }$

We know :

⇒ (a - b)(a + b) = a² - b²

Since :

$\rightarrow \bold{\dfrac{2}{(1)^2-(sin^2A)} }$

We know :

⇒ sin²A + cos²A = 1

Since :

$\rightarrow \bold{\dfrac{2}{cos^2A} }$

We know :

⇒ cos A = 1/sec A

Hence :

$\rightarrow \bold{2sec^2A}$

∴ L.H.S = R.H.S, Hence proved !

Answer 2

Explanation:

■ GIVEN THAT:-

$\frac{1}{1 - \sin \: A } + \frac{1}{1 + \sin \: A } = 2\sec ^{2} A$

■ SOLUTION:-

LHS:-

$\frac{1}{1 - \sin \: A } + \frac{1}{ \sin \: A }$

$\frac{1 + \sin \: A + 1 - \sin \: A }{(1 - \sin \: A)(1 + \sin \: A) }$

$\frac{2}{1 - \sin^{2} A }$

$\frac{2}{ \cos^{2} A }$

$= 2 \sec^{2} A$

LHS=RHS

HENCE PROVED!