Question

Plz give me a answer

Answer 1

`sec^2 \alpha -tan^2 \alpha =1, sec^2 \alpha =1+tan ^2 \alpha =1+( (3)/(4) ^2=1+ (9)/(16) = (16+9)/(16) = (25)/(16)`,

`as tan \alpha is positive so \alpha lies in 1st or 3rd quadrant only. But \alpha does not lie in 1st`
`so it lies in 3rd quadrant. sec \alpha =- \sqrt{ (25)/(16) } =- (5)/(4) , cos \alpha =- (4)/(5)`

`tan \alpha = (3)/(4), (sin \alpha )/(cos \alpha ) = (3)/(4) , (sin \alpha )/(- (4)/(5) ) = (3)/(4) ,`

`sin \alpha = (3)/(4) * (-4)/(5) =- (3)/(5)`

`sec \beta = (13)/(5) ,cos \beta = (5)/(13) cos \beta is positive,it lies in 4th quadrant(not lies in 1st )`

`sin \beta =- √(1-cos^2 \beta ) =- \sqrt{1-( (5)/(13))^2 } =- \sqrt{ (169-25)/(169) } =- \sqrt{ (144)/(169) } =- (12)/(13)`
sin(α+β)=sinαcosβ+cosαsinβ

`sin( \alpha + \beta )=- (3)/(5) * (5)/(13) +( (-4)/(5) )*( (-12)/(13) ) = (-15)/(65) + (48)/(65) = (-15+48)/(65) = (33)/(65)`