Composition of Trigonometric Functions and Inverse Trigonometric Functions
if a=cos 2 B+...
Question
if a = cos 2B + cos 2A. , b= cos 2B - cos 2 A
c= sin 2A+ sin 2B, d= sin = sin2A-sin2B
Then which of the first is true.
(A) a/b= cot(A+B) cot(A-B)
(B) c/d= tan(A+B)/ tan(A-B)
(C ) b/ c= tan(A-B)
(D) none of these
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Solution
At first we need to know about certain equation they are sin A + sin B = 2 sin [ (A + B) / 2 ] cos [ (A - B) / 2 ]
sin A - sin B = 2 cos [ (A + B) / 2 ] sin [ (A - B) / 2 ]
cos A + cos B = 2 cos [ (A + B) / 2 ] cos [ (A - B) / 2 ]
cos A - cos B = - 2 sin [ (A + B) / 2 ] sin [ (A - B) / 2 ]
So first look at the 1st option (a/b) So (a/b)=[cos2b+cos2a]/[cos2b-cos2a] So applying the equation given above (a/b)=[2cos(2a+2b)/2 cos(2a-2b)/2]/[2sin(2a+2b)/2 sin(2b-2a)/2]
=[cos(a+b)cos(b-a)]/[[sin(a+b)sin(b-a)] So (a/b)=[cot(a+b)cot(a-b)] So option a is correct
Now next option c/d=[sin2a+sin2b]/[sin2a-sin2b] now applying the equation given above c/d=[2sin(2a+2b)/2 cos(2a-2b)/2]/[2cos(2a+2b)/2 sin(2a-2b)/2] Now simplifying and cancelling 2 c/d=[sin(a+b)cos(a-b)]/[cos(a+b)sin(a-b)] c/d=tan(a+b)cot(a-b) Ie.c/d=tan(a+b)/tan(a-b)
So option b is correct...
Now we look for 3rd option b/c=[cos2b-cos2a]/[sin2a+sin2b]
Applying the formulas b/c=[2sin(2a+2b)/2 sin(2b-2a)/2]/[2sin(2a+2b)/2 cos(2b-2a)/2]
Simplifying and cancelling those 2 we get c/d=(tan(b-a)) So our option c is not correct