यदि त्रिभुज ABC में, `angle C =90^(@)` हो, तो सिद्ध कीजिए कि - `"tan"(A)/(2)=sqrt((c-b)/(c+b))`.

Explanation

`because " " (c-b)/(c+b)=(sin C-sin B)/(sin C + sin B) " "` [ज्या नियम से]
`rArr " " (c-b)/(c+b)=(sin 90^(@)-sin B)/(sin 90^(@)+sin B) " "` [`because angle C = 90^(@)` दिया है]
`=(1-sin B)/(1+sin B) " " [because sin 90^(@)=1]`
`=(1-sin (90^(@)-A))/(1+sin (90^(@)-A))" " [because angle A + angle B = angle C = 90^(@)]`
`=(1-cos A)/(1+cos A) " " [because sin (90^(@)-A)=cos A]`
`=("2 sin"^(2)(A)/(2))/("2 cos"^(2)(A)/(2)) " " [because 1-cos theta = 2 sin^(2) theta//2` एवं `1+cos theta = 2 cos^(2)theta//2]`
`= tan^(2)A//2 " " [because sin theta//cos theta = tan theta]`
`rArr " " tan^(2)A//2=(c-b)/(c+b) rArr tan A//2 = sqrt((c-b)/(c+b)). " "` इति सिद्धम्