Question

Plz give me a answer

Answer 1

$sec^2 \alpha -tan^2 \alpha =1, sec^2 \alpha =1+tan ^2 \alpha =1+( \frac{3}{4} ^2=1+ \frac{9}{16} = \frac{16+9}{16} = \frac{25}{16}$,
$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{ \frac{25}{16} } =- \frac{5}{4} , cos \alpha =- \frac{4}{5}$
$tan \alpha = \frac{3}{4}, \frac{sin \alpha }{cos \alpha } = \frac{3}{4} , \frac{sin \alpha }{- \frac{4}{5} } = \frac{3}{4} ,$
$sin \alpha = \frac{3}{4} * \frac{-4}{5} =- \frac{3}{5}$
$sec \beta = \frac{13}{5} ,cos \beta = \frac{5}{13} cos \beta is positive,it lies in 4th quadrant(not lies in 1st )$
$sin \beta =- \sqrt{1-cos^2 \beta } =- \sqrt{1-( \frac{5}{13})^2 } =- \sqrt{ \frac{169-25}{169} } =- \sqrt{ \frac{144}{169} } =- \frac{12}{13}$
sin(α+β)=sinαcosβ+cosαsinβ
$sin( \alpha + \beta )=- \frac{3}{5} * \frac{5}{13} +( \frac{-4}{5} )*( \frac{-12}{13} ) = \frac{-15}{65} + \frac{48}{65} = \frac{-15+48}{65} = \frac{33}{65}$