Grade 11Trigonometry

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

Narinderpal

9 Years agoGrade 11

2 Answers

Aman

9 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)〗=(-627)/725

Shankar

6 Years ago

Given that tan(α2) and tan(β2) are the roots of equation 8x2−26x+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) =−627725⇒cos(α+β)=−627725Given that tan(α2) and tan(β2) are the roots of equation 8x2−26x+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) =−627725⇒cos(α+β)=−627