Given:
cosecA + cotA = a√b
Concept:
We'll use trigonometric identities to manipulate the given equation and express it in terms of sinA and cosA
Formula Used:
cosecA = 1/sinA,
cotA = cosA/sinA
sin2A + cos2A = 1
Calculation :
cosecA + cotA = a√b
1/sinA + cosA/sinA = a√b
(1 + cosA)/sinA = a√b
Now, we'll square both sides of the equation
{(1 + cosA)/sinA}2 = (a√b)2
sin2A(1+cosA)2=a2b
1−cos2A1+2cosA+cos2A=a2b __________Eqn 1
Subtracting -1 in both sides, we gets
1−cos2A1+2cosA+cos2A−1=a2b−1
1−cos2A1+2cosA+cos2A−1+cos2A=a2b−1
1−cos2A2cosA+2cos2A=a2b−1
= 1−cos2A2cosA(1+cosA)=a2b−1 ________Eqn 2
Adding +1in both sides to eqn1
1−cos2A1+2cosA+cos2A+1=a2b+1
1−cos2A1+2cosA+cos2A+1−cos2A=a2b+1
1−cos2A2+2cosA=a2b+1
= 1−cos2A2(1+cosA)=a2b+1 _______ Eqn3
Now, dividing eqn2 and eqn3
1−cos2A2(1+cosA)1−cos2A2cosA(1+cosA)=a2b+1a2b−1
cosA = a2b+1a2b−1
So, the answer is cosA