Question

please prove it class 11

Answer 1

Note: I am considering alpha as a and Beta as b.

$Given : tan \frac{a}{2} = \sqrt{\frac{1 - e}{1 + e}} tan \frac{b}{2}$

On squaring both sides, we get

$= > tan^2 \frac{a}{2} = (\frac{1 - e}{1 + e}) tan^2 \frac{b}{2}$

we know that tan^2(x/2) = (1 - cosx)/(1 + cosx)

$= > \frac{1 - cosa}{1 + cosa} = (\frac{1 - e}{1 + e})(\frac{1 - cosb}{1 + cosb})$

$= > (1 - cosa)(1 + e)(1 + cosb) = (1 + cosa)(1 - e)(1 - cosb)$

= > 1 + cosb + e + ecosb - cosa - cosacosb - ecosa - ecosacosb = 1 - cosb - e + ecosb + cosa - cosacosb - ecosa + ecosacosb

= > 2cosb - 2ecosacosb = 2cosa - 2e

= > 2(cosb - ecosacosb) = 2(cosa - e)

= > cosb - ecosacosb = cosa - e

= > cosb(1 - ecosa) = cosa - e

$= > cosb = \frac{cosa - e}{1 - ecosa}$

Hence proved.

Hope it helps!