Question

if sin beta+cos beta=m, sec beta+cosec beta=n, prove that n(m^2-1)=2m

Answer 1

consider Lhs
sec beta +cosec beta((sin beta +cos beta )^2 -1)
  sec can be written in terms of cos and cosec can be written in terms of sin

1/cos beta + 1/sin beta ( sin^2 beta + cos^2 beta + 2sin beta cos beta ) -1
on taking lcm
        cos beta    +    sin beta /cos beta sin beta    (1  + 2 sin beta cos beta   -1)  as cos^2 +sin^2 is one

now 1 - 1 gets cancelled  and also cos beta sin beta ets cancelled
 so we remain with

2cos beta + 2sin beta

rhs
2m = 2(sin beta + cos beta)
2sin beta + 2cos beta
hens prooved

Answer 2

see here b=beta
sinb+cosb=m
secb+cosecb=n  => 1/cosb +1/sinb
=> n(m^2-1)=2m
rhs
=> 1/cosb + 1/sinb{(sinb + cosb)^2-1}
      (sinb+cosb)/sinbcosb{sin^2b + cos^2b+2sinbcosb - 1}
       (sinb+cosb)/sinbcosb(1+2sinbcosb-1)
(sinb+cosb)/sinbcosb(2sinbcosb)
(sinb+cosb)2
2sinb+2cosb
2(sinb+cosb)
hence shown