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

Question

Aman · Answer

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 · Answer

G i v e n t h a t tan ( α 2 ) a n d tan ( β 2 ) a r e t h e r o o t s o f e q u a t i o n 8 x 2 − 26 x + 15 = 0 . S o tan ( α 2 ) + tan ( β 2 ) = − − 26 8 = 13 4 A n d , tan ( α 2 ) tan ( β 2 ) = 15 8 T h u s , tan ( α + β 2 ) = tan ( ( α 2 ) + ( β 2 ) ) = tan ( α 2 ) + tan ( β 2 ) 1 − tan ( α 2 ) tan ( β 2 ) = 13 4 1 − ( 15 8 ) = − 26 7 N o w , cos ( α + β ) = 1 − tan 2 ( α + β 2 ) 1 + tan 2 ( α + β 2 ) = 1 − ( 676 49 ) 1 + ( 676 49 ) = − 627 725 ⇒ c o s ( α + β ) = − 627 725 G i v e n t h a t tan ( α 2 ) a n d tan ( β 2 ) a r e t h e r o o t s o f e q u a t i o n 8 x 2 − 26 x + 15 = 0 . S o tan ( α 2 ) + tan ( β 2 ) = − − 26 8 = 13 4 A n d , tan ( α 2 ) tan ( β 2 ) = 15 8 T h u s , tan ( α + β 2 ) = tan ( ( α 2 ) + ( β 2 ) ) = tan ( α 2 ) + tan ( β 2 ) 1 − tan ( α 2 ) tan ( β 2 ) = 13 4 1 − ( 15 8 ) = − 26 7 N o w , cos ( α + β ) = 1 − tan 2 ( α + β 2 ) 1 + tan 2 ( α + β 2 ) = 1 − ( 676 49 ) 1 + ( 676 49 ) = − 627 725 ⇒ c o s ( α + β ) = − 627

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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)=?

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