Question

SinA+CosA=P and SecA+CosecA=q
Show that p²q-2pq-q=0

Answer 1

Answer:

Step-by-step explanation:

Correct Question :-

SinA+CosA=P and SecA+CosecA=q

Show that p²q-2p-q=0

Given,

sinA + cosA = p

secA + cscA = q

To Prove :-

p²q-2p-q=0

Formula Required :-

sin²A + cos²A = 1

(a + b)² = a² + b² + 2ab

Solution :-

Taking L.H.S :-

p²q - 2pq - q

Substituting the values of both 'p' and 'q' :-

= (sinA + cosA)²(secA + cscA) - 2(sinA + cosA) - (secA + cscA)

= (sin²A + cos²A + 2sinAcosA)(secA + cscA) - 2[sinA + cosA ]- secA - cscA

[∴ (a + b)² = a² + b² + 2ab]

= (1 + 2sinAcosA)(secA + cscA) - 2sinA - 2cosA - secA - cscA

[ ∴ sin²A + cos²A = 1]

= [1(secA + cscA) + 2sinAcosA(secA + cscA)] - 2sinA - 2cosA - secA - cscA

= [secA + cscA + 2sinAcosAsecA + 2sinAcosAcscA] - 2sinA - 2cosA - secA - cscA

= [secA + cscA + 2sinAcosA×1/cosA + 2sinAcosA×1/sinA] - 2sinA - 2cosA - secA - cscA

= [secA + cscA + 2sinA + 2cosA] - 2sinA - 2cosA - secA - cscA

= secA + cscA + 2sinA + 2cosA - 2sinA - 2cosA - secA - cscA

= secA - secA + cscA - cscA + 2sinA - 2sinA + 2cosA - 2cosA

= 0

= R.H.S

Hence Proved.