यदि ` sin A = 3/5 , cos B = (-12)/13 ` जहाँ ` A in [ 0, (pi)/2 , B in ] pi , (3pi)/2` , तो निम्नलिखित के मान ज्ञात कीजिए ।
i) sin ( A − B ) ii) cos ( A − B ) iii) tan ( A + B )

Explanation

` sin A = 3/5 `
` :. cos^(2) A = 1 - sin^(2) A `
`= 1 - (3/5)^(2) = 16/25`
` rArr cos A = 4/5 `
और ` tan A = (sin A)/(cos A) = (3/5)/(4/5)= 3/4 `
पुनः ` cos B = (-12)/13 , B ` तृतीय चतुर्थाश में है ।
` :. sin^(2) B = 1 - cos^(2) B`
` =1 - ((-12)/13)^(2) = 25/169`
` rArr sin B = (-5)/13`
(तृतीय चतुर्थाश में ऋणात्मक होता है )
` :. tan B = ( sin B )/(cosB)`
` = ((-5)/(13))/((-12)/13)= 5/12`
(i) ` sin (A-B) = sin A cos B - cos A sin B`
` = 3/5 ((-12)/13)- 4/5 ((-5)/13)`
` = (-36 + 20)/65 = (-16)/65`
(ii) ` cos ( A-B) = cos A cos B + sin A sin B`
` = 4/5 ((-12)/13 ) + 3/5 ((-5)/13)`
` = (-48 -15)/65 = (-63)/65`
(iii) ` tan ( A + B) = ( tan A + tan B ) /( 1 - tan A tan B) `
` (m/(m+1)+1/(2m+1))/(1- m/(m+1)*1/(2m+1))`
` ((m+1) (2m+1))/(((m+1)(2m+1)-m)/((m+1)(2m+1)))`
` ( 2m^(2) + m + m +1)/(2m^(2)+ 2m+ m +1 -m)`
` = (2m^(2) + 2m +1)/(2m^(2) + 2m+1)`
` = 1 = tan. pi/4`
` = 1= tan. pi/4`
` :. A + B = pi/4`