Question

If sinθ+cosθ=p and secθ+cosecθ=q, show that q(p²-1)=2p.Prove it by using trigonometric identities.

Answer 1

GIVEN :

If sinθ+cosθ=p and secθ+cosecθ=q, show that q(p²-1)=2p

LHS = q(p²-1)

= (secθ+cosecθ)[(sinθ+cosθ)²-1]

=(1/cosθ+1/sin θ) (sin²θ+2 sinθ cosθ+cos²θ-1)

[cosecθ = 1/sinθ , secθ = 1/cosθ, (a+b)² = a² + b² + 2ab]

=(1/cosθ+1/sin θ) (sin²θ + cos²θ +2 sinθ cosθ-1)

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

[sin²θ+cos²θ=1]

={(sinθ+cosθ)/sinθ cosθ}(2 sinθ cosθ )

=2(sinθ+cosθ)

LHS =2(sinθ+cosθ) = 2p = RHS

HOPE THIS ANSWER WILL HELP YOU.

Answer 2

Hi ,

Here I am using 'A ' instead of theta.

sinA + cosA = p ----( 1 )

do the square of equation ( 1 ) , we get

( sinA + cosA )² = p²

sin²A + cos²A + 2sinAcosA = p²

1 + 2sinAcosA = p² ---( 2 )

secA + cosecA = q

1/cosA + 1/sinA = q

( sinA + cosA )/sinAcosA = q

p/sinAcosA = q ----( 3 ) [ from ( 1 ) ]

Now ,

LHS = q ( p² - 1 )

= ( p/sinAcosA )[ 1 + 2sinAcosA - 1 ]

{ from ( 3 ) and ( 2 ) }

= ( p/sinAcosA ) × ( 2sinAcosA )

after cancellation, we get

= 2p

= RHS

I hope this helps you.

: )