If `k = tan25^@`then`(K +1)/(k-1) + ( k-1)/(k+1)`

If t a ntheta_1t a ntheta_2=k , then ("cos"(theta_1-theta_2))/("cos"(theta_1+theta_2))= a. (1+k)/(1-k) b. (1-k)/(1+k) c. (k+1)/(k-1) d. (k-1)/(k+1)

If tantheta_1=kcottheta_2, then find cos(theta_1-theta_2)/cos(theta_1+theta_2)= ................ A) (1+k)/(1-k) B) (1-k)/(1+k) C) (k+1)/(k-1) D) (k-1)/(k+1)

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

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

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

If A+B+C=pi&sin(A+(C)/(2))=k(sin C)/(2), then (tan A)/(2)(tan B)/(2)= a.(k-1)/(k+1) b.(k+1)/(k-1) c.(k)/(k+1) d.(k+1)/(k)

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 s e ctheta+t a ntheta=k ,\ costheta= a. (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 tan A=k tan B show that sin(A+B)=((k+1)/(k-1))sin(A-B)