Question

If
$\frac{ \sin(a) }{1 + \sin(a) } + \frac{ \sin(b) }{1 + \sin(b) } = 2$
; prove that
${ \sec(a) }^{2} (1 - \sin(a)) + { \sec(b) }^{2} (1 - \sin(b)) = 0$
.​

Answer 1

Given:

$\dfrac{sin A}{1 + sin A} + \dfrac{sin B}{1 + sin B} = 2$

To Prove:

sec² A (1 - sin A) + sec² B (1 - sin B) = 0

Proof:

$\dfrac{sin A (1 + sin B) + sin B (1 + sin A)}{(1 + sin A) (1 + sin B)} = 2$

sin A + sin A sin B + sin B + sin A sin B = 2 (1 + sin B + sin A + sin A sin B)

sin A + sin B + 2 sin A sin B = 2 sin A + 2 sin B + 2 sin A sin B + 2

(2 sin A - sin A) + (2 sin B - sin B) + 2 = 0

sin A + sin B + 2 = 0  →→→→→→→→→ (Equation 1)

sec² A (1 - sin A) + sec² B (1 - sin B) = 0

LHS = sec² A (1 - sin A) + sec² B (1 - sin B)

$\dfrac{1}{cos^{2} A} (1 - sin A) + \dfrac{1}{cos^{2} B} (1 - sin B)$

$\dfrac{1 - sin A}{cos^{2} A} + \dfrac{1 - sin B}{cos^{2} B}$

$\\\dfrac{1 - sin A}{(1 + sin A) (1 - sin A)} + \dfrac{1 - sin B}{(1 + sin B) (1 - sin B)} \\\dfrac{1}{1 + sin A} + \dfrac{1}{1 + sin B} \\\dfrac{1 + sin B + 1 + sin A}{(1 + sin A) (1 + sin B)}$

$\dfrac{sin A + sin B + 2}{(1 + sin A) (1 + sin B)}$

From Equation 1, we have:

$\dfrac{0}{(1 + sin A) (1 + sin B)} = 0 = RHS$

Hence Proved.

Answer 2

Answer:

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