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If tan Alpha /2and tan beta/2aee the roots of es8x2-26x+15,thencos(Alpha+beta)=?

If tan Alpha /2and tan beta/2aee the roots of es8x2-26x+15,thencos(Alpha+beta)=?

Grade:11

2 Answers

Aman
15 Points
5 years ago
8x^2-26x+15=0
tan⁡〖α/2〗+tan⁡〖β/2〗=-b/a=26/8=13/4
tan⁡〖α/2〗  tan⁡〖β/2〗=c/a=15/8
tan⁡〖(α+β)/2〗=(tan⁡〖α/2〗+tan⁡〖β/2〗)/(1-tan⁡〖α/2〗  tan⁡〖β/2〗 )
tan⁡〖(α+β)/2〗=(13/4)/(1-15/8)=-26/7
cos⁡〖(α+β)〗=(1-tan^2⁡〖(α+β)/2〗)/(1+tan^2⁡〖(α+β)/2〗 )
substituting the values, we get
cos⁡〖(α+β)〗=(1-(-26/7)^2)/(1+(-26/7)^2 )
cos⁡〖(α+β)〗=(49-676)/(49+676)
cos⁡〖(α+β)〗=(49-676)/(49+676)
cos⁡〖(α+β)〗=(-627)/725
 
Shankar
20 Points
2 years ago
Given that tan(α2) and tan(β2) are the roots of equation 8x226x+15=0. So tan(α2)+tan(β2)=268=134And,  tan(α2)tan(β2)=158Thus, tan(α+β2)= tan((α2)+(β2))                     = tan(α2)+tan(β2)1 tan(α2)tan(β2)                     =1341(158)                     =267Now, cos(α+β)=1 tan2(α+β2)1+ tan2(α+β2)                 =1(67649)1+(67649)                 =627725cos(α+β)=627725Given that tan(α2) and tan(β2) are the roots of equation 8x226x+15=0. So tan(α2)+tan(β2)=268=134And,  tan(α2)tan(β2)=158Thus, tan(α+β2)= tan((α2)+(β2))                     = tan(α2)+tan(β2)1 tan(α2)tan(β2)                     =1341(158)                     =267Now, cos(α+β)=1 tan2(α+β2)1+ tan2(α+β2)                 =1(67649)1+(67649)                 =627725cos(α+β)=627
 

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