यदि `Delta ABC` में `angle C = 90^(@)` हो, तो सिद्ध कीजिए कि - `a(1+"tan"(B)/(2))=(b+c)(1-"tan"(B)/(2))`.

Explanation

चूँकि `" " (b+c)/(a)=(sin B + sin C)/(sin A)`
`=(sin B + sin 90^(@))/(sin A) " " [because angle C = 90^(@)]`
`=(sin B+1)/(sin A)" " [because sin 90^(@)=1]`
`=(1+sin B)/(sin((pi)/(2)-B))" " [because angle A + angle B = angle C = 90^(@)]`
`("sin"^(2)(B)/(2)+"cos"^(2)(B)/(2)+"2 sin"(B)/(2)"cos"(B)/(2))/(cos B)`
`rArr " " (b+c)/(a)=(("sin"(B)/(2)+"cos"(B)/(2))^(2))/(("cos"^(2)(B)/(2)-"sin"^(2)(B)/(2)))`
`=(("sin"(B)/(2)+"cos"(B)/(2))^(2))/(("cos"(B)/(2)-"sin"(B)/(2))("cos"(B)/(2)+"sin"(B)/(2)))`
`=(("sin"(B)/(2)+"cos"(B)/(2)))/(("cos"(B)/(2)-"sin"(B)/(2)))`
`=("tan"(B)/(2)+1)/(1-"tan"(B)/(2))`
`rArr " " a(1+"tan"(B)/(2))=(b+c)(1-"tan"(B)/(2))." "` इति सिद्धम्