Math, asked by lakshayjain9098, 1 year ago

if tanθ=2mn/m2-n2 then find the value of sinθ and cosθ

Answers

Answered by adarsh08
1

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 }

Answered by igamer2131
0

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

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