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if sin x=3/5 , cos y= -12/13 and x,y both lie in the second quadrant find the value of:- (a) cos(x-y) (b) tan(x+y) if sin x=3/5 , cos y= -12/13 and x,y both lie in the second quadrant find the value of:- (a) cos(x-y)(b) tan(x+y)
We, are given information that both sin(x) and cos(y) lie in second quadrant.By that we can say that sin will be positive and cos will be negative.sin(x)=3/5cos(x)=(1 – sin2(x))1/2cos(x)=+(or) – 4/5Since,(x) is in 2nd quadrant, cos(x) will – 4/5cos(x – y) = cos(x)cos(y) + sin(x)sin(y)Similarly, sin(y)=5/13.By, applying the above formula we get:cos(x-y) = (-)4/5*(-)12/13 + 3/5*5/13 = 63/65 For, the second part we would be using the given below formula:tan(x+y) = (tan(x) + tan(y)) ÷ (1 – tan(x)tan(y))tan(x)=3/5 & tan(y)=5/12tan(x+y) = [3/5 + 5/12]÷[1 – 3/5×5/12] = 61/45
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