1.

If `angleA=90^(@)` in the `DeltaABC` , then `tan^(-1)((c)/(a+b))+tan^(-1)((b)/(a+c))` is equal to

Answer» Correct Answer - C
`becauseDeltaABC` right angled at A.
`thereforea^(2)=b^(2)+c^(2)`
Now , `tan^(-1)((c)/(a+b))+tan^(-1)((b)/(a+c))`
`=tan^(-1)[((c)/(a+b)+(b)/(a+c))/(1-((c)/(a+b))((b)/(a+c)))]`
`=tan^(-1)[(ac+c^(2)+ab+b^(2))/(a^(2)+ac+ab+bc-bc)]`
`=tan^(-1)[(a^(2)+ac+ab)/(a^(2)+ac+ab)]` [using Eq . (i)]
`=tan^(-1)(1)=(pi)/(4)`


Discussion

No Comment Found

Related InterviewSolutions