Question

4) If sec-m+n÷2√mn find sin​

Answer 1

Answer:

May this answer helped you

Step-by-step explanation:

Solution:

1/m[(m + n) + 1/(m + n)]

= 1/m [{(m + n)2 + 1}/ (m + n)]

= 1/m [(m2 + n2 + 2mn + 1)/ (m + n)]

Substituting sec θ = m and tan θ = n,

= 1/sec θ[(sec2θ + tan2θ + 2 sec θ tan θ + 1)/ (sec θ + tan θ)]

= 1/sec θ[(sec2θ + sec2θ + 2 sec θ tan θ)/ (sec θ + tan θ)]

= 1/sec θ[(2 sec2θ + 2 sec θ tan θ)/ (sec θ + tan θ)]

= 1/sec θ[2 sec θ(sec θ + tan θ)/ (sec θ + tan θ)]

= 2 sec θ/sec θ

= 2