If `A+B+C=pi\ &sin(A+C/2)=ksinC/2, t h e ntanA/2tanB/2=` a.`(k-1)/(k+1)` b. `(k+1)/(k-1)` c.`k/(k+1)` d. `(k+1)/k`

A+B=C and tan A=k tan B, then prove that sin(A-B)=(k-1)/(k+1)sin C

If sec theta+tan theta=k,cos theta=(k^(2)+1)/(2k)b(2k)/(k^(2)+1)c*(k)/(k^(2)+1)d*(k)/(k^(2)-1)

If tan theta_(1)tan theta_(2)=k, then (cos(theta_(1)-theta_(2)))/(cos(theta_(1)+theta_(2)))=(1+k)/(1-k) b.(1-k)/(1+k) c.(k+1)/(k-1) d.(k-1)/(k+1)

If an angle alpha is divided into two parts A&B such that A-B=x and tan A:tan B=k:1 Then the value of sin x is (A) (k+1)/(k-1)sin alpha (k)/(k+1)sin alpha (C) (k-1)/(k+1)sin alpha (D) tan alpha

If an angle 'r' is divided in two parts A and B such that A-B=x and tan A: tan B=K:1 then the value of sin x is 1) (k+1)/(k-1)sin r, 2) (k)/(k+1)sin r 3) (k-1)/(k+1)sin r 4) None

If (tan3A)/(tan A)=k(k!=1) then which of the following is not true? (a) (cos A)/(cos3A)=(k-1)/(2)(b)(sin3A)/(sin A)=(2k)/(k-1)(c)(cot3A)/(cot A)=(1)/(k)(d) none of these

Lt_(nrarroo) sum_(r=1)^n (2r)^k/n^(k+1),k!=-1 , is equal to (A) 2^k/(k-1) (B) 2^k/k (C) 1/(k+1) (D) 2^k/(k+1)

The values of k for which the equation sin^(4)x+cos^(4)x+sin2x+k=0 possess solution (A)k in[-(3)/(2),1] (B) k in[0,(1)/(2)] (C) k in[-(3)/(2),(1)/(2)] (D) k in[(1)/(2),1]

If the points (k,2k),(3k,3k) and (3,1) are collinear,then k(1)/(3)(b)-(1)/(3)(c)(2)/(3)(d)-(2)/(3)

Consider a family of circles which are passing through the point (-1,1) and are tangent to the x-axis.If (h,k) are the coordinates of the center of the circles,then the set of values of k is given by the interval.k>=(1)/(2)(b)-(1)/(2)<=k<=(1)/(2)k<=(1)/(2)(d)0