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प्रश्न
If sin A = `3/5`, 0 < A < `pi/2` and cos B = `(-12)/13`, π < B < `(3pi)/2`, find the values of the following:
- cos(A + B)
- sin(A – B)
- tan(A – B)
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उत्तर
Given that sin A = `3/5`, 0 < A < `pi/2` (i.e., A lies in first quadrant)
Since A lies in first quadrant cos A is positive.

cos A = `"Adjacent side"/"Hypotenuse" = 4/5`
Tan A = `3/4`
AB = `sqrt(5^2 - 3^2)` = 4
Also given that cos B = `(-12)/13`, π < B < `(3pi)/2` (i.e., B lies in third quadrant)
Now sin B lies in third quadrant. sin B is negative.

CA = `sqrt(13^2 - 12^2)` = 5
sin B = `(- "Opposite side")/("Hypotenuse") = (-5)/13`
tan B = `(- "Opposite side")/("Adjacent") = (5)/12` .... [B lies in 3rd quadrant. tan B is positive.]
(i) cos(A + B) = cos A cos B – sin A sin B
`= 4/5((-12)/13) - 3/5 xx ((-5)/13)`
`= (-48)/65 + 15/65 = (-33)/65`
(ii) sin(A – B) = sin A cos B – cos A sin B
`= 3/5((-12)/13) - 4/5 xx ((-5)/13)`
`= (-36)/65 + 20/65 = (-16)/65`
(iii) tan(A – B)
tan(A – B) = `(tan "A" - tan "B")/(1 + tan "A" tan "B")`
`= (3/4 - (5/12))/(1 + 3/4 xx (5/12))`
`= (3/4 - 5/12)/(1 + 3/4 xx 5/12)`
`= ((9-5)/12)/(1 + 5/(4 xx 4))`
`= (4/12)/(21/16)`
`= 4/12 xx 16/21`
`= (4 xx 4)/(3 xx 21)`
`= 16/63`
