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If sinx+ siny=a and cosx +cosy=b, then tan^2(x+y/2) tan^2(x-y/2) =?
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If sinx+ siny=a and cosx +cosy=b, then tan^2(x+y/2) tan^2(x-y/2) =?
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Given:-
eqn 1 : sin x + sin y =a
2.sin x+y/2 cos x-y/2 = a
eqn 2 : cos x + cos y = b
2.cos x+y/2 cos x-y/2 =b
DIVIDING eqn1 by eqn2 :-
a/b= {2.sin x+y/2 cos x-y/2 } /
{2.cos x+y/2 cos x-y/2 }
now cancelling the similar one
{sin x+y/2}/ {cos x+y/2 }
tan (x+y/2) = a/b
ON SQUARING egn1 and eqn2 , WE GET :-
(eqn1)^2 + (eqn2)^2 = sin^2x +sin^2y + 2. sin x. sin y + cos^2x+cos^2y+ 2 cos x.cos y =a^2+b^2
APPLYING IDENTITIES AND TAKING 2AS COMMON :-
1+1+2(cos x. cos y + sin x.sin y)=a^2+b^2
2[cos(x-y)] = a^2 +b^2 - 2
cos(x-y) = a^2 +b^2 -2/2
tan^2(x-y/2) = 1-cos(x-y) / 1+cos(x+y)
= (1- a^2 + b^2 - 2 /2) / (1+a^2+b^2-2/2)
= 4-a^2-b^2/a^2+b^2
tan^2 (x+y/2) + tan^2 (x-y/2)
= a^2/b^2 + 4-a^2-b^2/ a^2+b^2
= {a^2(a^2+b^2) + b^2(4-a^2-b^2)}/
{ b^2 ( a^2+b^2) }
={a^4+a^2b^2+4b^2-a^2b^2-b^4}/
{ a^2 b^2 + b^4 }
CANCELLING +a^2b^2 with -a^2b^2 , WE GET :-
(a^4-b^4+4b^2)/ (a^2b^2+b^4

this is your answer dude:-
{a^4 - b^4 + 4b^2}/ { a^2b^2 + b^4}
Community Answer
If sinx+ siny=a and cosx +cosy=b, then tan^2(x+y/2) tan^2(x-y/2) =?
Solution

To solve the problem, we will use the trigonometric identities of sin and cos to express tan^2(x y/2) and tan^2(x-y/2) in terms of a and b.

Expressing tan^2(x y/2) in terms of a and b

We know that:
tan(x y/2) = sin(x y)/(cos(x) + cos(y))

Squaring both sides, we get:
tan^2(x y/2) = sin^2(x y)/(cos^2(x) + 2cos(x)cos(y) + cos^2(y))

Using the identity sin^2(x) + cos^2(x) = 1, we can simplify the denominator to:
tan^2(x y/2) = sin^2(x y)/(1 + 2cos(x)cos(y))

We are given that sin(x)sin(y) = a and cos(x)cos(y) = b. Using these values, we can substitute in the expression to get:
tan^2(x y/2) = (a/b)^2 / (1 + 2b)

Expressing tan^2(x-y/2) in terms of a and b

We know that:
tan(x-y/2) = sin(x-y)/(cos(x) + cos(y))

Using the identity sin(x-y) = sin(x)cos(y) - cos(x)sin(y), we can simplify the numerator to:
tan(x-y/2) = (sin(x)cos(y) - cos(x)sin(y))/(cos(x) + cos(y))

Dividing the numerator and denominator by cos(x)cos(y), we get:
tan(x-y/2) = (a/b - 1)/(1 + b)

Squaring both sides, we get:
tan^2(x-y/2) = (a^2 + b^2 - 2ab)/(b^2 + 2b + 1)

Final Solution

We can now substitute the expressions for tan^2(x y/2) and tan^2(x-y/2) in the required expression and simplify to get:
tan^2(x y/2) tan^2(x-y/2) = [(a/b)^2 / (1 + 2b)][(a^2 + b^2 - 2ab)/(b^2 + 2b + 1)]
= a^2 / (b^2 + 2b + 1)

Therefore, tan^2(x y/2) tan^2(x-y/2) = a^2 / (b^2 + 2b + 1)
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If sinx+ siny=a and cosx +cosy=b, then tan^2(x+y/2) tan^2(x-y/2) =?
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