`4-2a+2a=2a-a^2`

If sinalpha+sinbeta=a and cosalpha+cosbeta=b , prove that tan((alpha-beta)/2)=+-sqrt((4-a^2-b^2)/(a^2+b^2)) .

If sinalpha+sinbeta=a and cosalpha+cosbeta=b , prove that tan((alpha-beta)/2)=+-sqrt((4-a^2-b^2)/(a^2+b^2)) .

Solve the following quadratic equations by factorization method: 4x^2-2(a^2+b^2)x+a^2b^2=0

If sin^-1 (x/a) +sin^-1 (y/b) = sin^-1 (c^2/(ab)) then prove that b^2x^2 +2xysqrt(a^2b^2-c^2) = c^4-a^2y^2-2x^2y^2

The value of the determinant [(1^2, 2^2, 3^2, 4^2) ,(2^2, 3^2 ,4^2,5^2) ,(3^2, 4^2,5^2 ,6^2) ,(4^2, 5^2, 6^2, 7^2)] is equal to 1 b. 0 c. 2 d. 3

The locus of a point, from where the tangents to the rectangular hyperbola x^2-y^2=a^2 contain an angle of 45^0 , is (a) (x^2+y^2)^2+a^2(x^2-y^2)=4a^2 (b) 2(x^2+y^2)^2+4a^2(x^2-y^(2))=4a^2 (c) (x^2+y^2)^2+4a^2(x^2-y^2)=4a^2 (d) (x^2+y^2)+a^2(x^2-y^(2))=a^4

The locus of a point, from where the tangents to the rectangular hyperbola x^2-y^2=a^2 contain an angle of 45^0 , is (a) (x^2+y^2)^2+a^2(x^2-y^2)=4a^2 (b) 2(x^2+y^2)^2+4a^2(x^2-y^(2))=4a^2 (c) (x^2+y^2)^2+4a^2(x^2-y^2)=4a^2 (d) (x^2+y^2)+a^2(x^2-y^(2))=a^4

If the curves x^(2) + 4y^(2) = 4, x^(2) + a^(2) y^(2) = a^(2) for suitable value of a cut on four concylclic points, the equation of the circle passing through the four points, is

In a triangle ABC, let 2a^2+4b^2+ c^2= 2a(2b + c) , then which of the following holds good?

Write the sum of the series 1^2-2^2+3^2-4^2+5^2-6^2...........+(2n-1)^2-(2n)^2dot