Question

1/1-sinθ+1/1+sinθ=2sec²θ, Prove it?

Answer 1

LHS = $\frac{1}{1-sin\theta}+\frac{1}{1+sin\theta}$

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

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

[ We know, (a - b)(a + b) = a² - b² , so, (1 - sinθ)(1 + sinθ) = 1 - sin²θ]

= $\frac{2}{cos^2\theta}$

[We know, sin²α + cos²α = 1 so,1 - sin²θ = cos²θ]

= $2sec^2\theta$ = RHS [ as you know, secΦ = 1/cosΦ so, sec²θ= 1/cos²θ]

hence proved

Answer 2

Here I am writing theta as A.

$Given : \frac{1}{1 - sinA} + \frac{1}{1 + sinA}$

$= \ \textgreater \ (\frac{1}{1 - sinA} * \frac{1 + sinA}{ 1 + sinA }) + (\frac{1}{sinA} * \frac{1 - sinA}{1 - sinA} )$

$= \ \textgreater \ \frac{1 + sinA}{(1 - sinA)(1 + sinA)} + \frac{1 - sinA}{(1 + sinA)(1 - sinA)}$

$= \ \textgreater \ \frac{(1 + sinA) + (1 - sinA)}{(1 + sinA)(1 - sinA)}$

$= \ \textgreater \ \frac{1 + sinA + 1 - sinA}{(1 +sin A)(1 - sinA)}$

$= \ \textgreater \ \frac{2}{(1 + sinA)(1 - sinA)}$

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

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

$= \ \textgreater \ 2sec^2A$


Hope it helps!