Question

If Sin θ+ Cos θ = m and Sec θ + Cosec θ = n then show that n(m² – 1) = 2m.

Answer 1

GIVEN:
Sin θ+  Cos θ = m…………….(1)
Sec θ + Cosec θ = n………….(2)

LHS = n(m² – 1)
LHS= (Sec θ + Cosec θ) { (Sin θ+  Cos θ )² - 1}

[From eq 1 & 2]

LHS = (1/cosθ + 1/sinθ ) {sin²θ + cos ²θ + 2 sinθ cosθ - 1}

[ Sec θ = 1/cosθ , Cosec θ = 1/sinθ , (a+b)² = a² +b²+2ab]

= (sinθ + cosθ / cosθ sinθ) ( 1 +  2 sinθ cosθ - 1)

[ sin² θ + cos²θ = 1]

= (sinθ + cosθ / cosθ sinθ) (2 sinθ cosθ )
LHS = 2(sinθ + cosθ) = 2m = RHS

[ From eq 1]

LHS = RHS

HOPE THIS WILL HELP YOU...

Answer 2

Hey..!!!

__________
__________
Given,

Sin θ+ Cos θ = m
Sec θ + Cosec θ = n

LHS = n ( m^2 -1)

= (secθ + cosecθ ) [ ( sinθ + cosθ )^2 -1]

= (secθ + cosecθ ) [ sin^2θ + cos^2θ + 2 sinθ cosθ -1]

= (secθ + cosecθ ) [ 1+2 sinθcosθ -1]

= ( secθ + cosecθ ) [ 2 sinθ cosθ ]

= ( 1/cosθ + 1/ sinθ ) [ 2 sinθ cosθ ]

= ( Sinθ + cosθ / sinθ cosθ ) [ 2 sinθ cosθ ]

= 2 ( sinθ + cosθ )

= 2m = RHS proved Ans.

LHS = RHS

________________
________________

I Hope it's help you...!!!

please tick the brainliest answer.