if tanθ=2mn/m2-n2 then find the value of sinθ and cosθ
Answers
Answer:
Step-by-step explanation:
tanθ = 2mn \ 2m - 2n
tanθ = 2mn \ 2(m-n)
tanθ = mn \ m-n
We know that,
sinθ = opp \ hyp
cosθ = adj \ hyp
tanθ = opp\ adj
from the ques,
tanθ = mn \ m-n = opp\ adj
So, opp = mn
adj = m-n
( hyp )^2 = (opp)^2 - (adj)^2
= ( mn )^2 - ( m-n )^2
hyp = whole under root { (mn)^2 - (m-n)^2 }
sinθ = mn \ whole under root { (mn)^2 - (m-n)^2 }
cosθ = m-n \ whole under root { (mn)^2 - (m-n)^2 }
Answer:
(sinx*cosx/sinx)cosx = 1
Step-by-step explanation:
sinx=2mn/(m^2+n^2)
cot = cos \div sin
(sinx*cotx)/cosx=1
LHS
(sinx*cosx/sinx)cosx
= 1