Question

if sin a +cos a=p, sec a+cosec a=q then show that
q(p2-1)=2p

Answer 1

P= sina + cosa => p^2 = (sina +cosa)^2= sin^2 a + cos^2 a + 2sinacosa = 1+ 2sinacosa now p^2-1= 2sinacosa now q(p^2-1) = (seca + coseca)2sinacosa = [1/cosa + 1/sina ] 2sinacosa = (sina + cosa)2 =2(sina+ cosa) = 2 p = RHS proved

Answer 2

Answer:

Step-by-step explanation:

Cosec A + Cot A = P

= 1/Sin A + Cos A / Sin A = P

= 1+Cos A/Sin A = p

=> SQUARING ON BOTH SIDES

= (1+COS A)²/(SINA)²= P²

= (1+ Cos A)² / (Sin A)² = P²

= (1+cos A)² = (p²)[(Sin A)²]

= (1+ cos A) ² = (p²) [(1-cos²A)]

= (1+ cos A) ² = (p²) [ (1+cosA)(1-cos A) ]

= (1+cos A)² ÷ (1+ cos A) = (p²)[(1-cos A)]

= 1+cos A = (p²)[1-cos A]

= 1+cos A ÷ 1-cos A = p²

= Here, Using (a+b/a-b=c+d/c-d). This is known as Componendo and dividendo

According to the Question statement!

1+cos A ÷ 1-cos A = p²

Then,

(1+cos A) + (1 - cos A ) ÷ (1+ Sin A - (1-sin A) = p²+1 / p²-1

= 2/2cos = p²+1/p²-1

= 1/cos = p²+1/p²-1

= Sec = p²+1 /p²-1

We know that Cos A = 1/sec A

Then,

Cos A = p²-1 /p²+1