Question

if sec theta + tan theta = p, prove that sinq =p²-1/p+1​

Answer 1

Answer:

$Given that sec A + tan A = p.Square both sides to getsec^2 A +2 sec A tan A + tan^2 A = p^2(1 + tan^2 A) +2 sec A tan A + tan^2 A = p^2, ortan^2 A +2 sec A tan A + tan^2 A = p^2 - 1, or2tan^2 A +2 sec A tan A = p^2 - 1, or2 tan A (tan A + sec A) = p^2 - 1, or2 tan A*p = p^2 - 1, ortan A = (p^2 - 1)/2pConsider a right angled triangle whose altitude is (p^2 - 1) and the base is 2p. The the hypotenuse = [(p^2 - 1)^2 + (2p)^2]^0.5= [p^4–2p^2+1+4p^2]^0.5= [p^4+2p^2+1]^0.5= (p^2+1)Hence$

Answer 2

Correct Question

• if sec theta + tan theta = p, prove that sin theta =p²-1/p²+1

Given:

• p = sec θ + tan θ

To Prove:

• sin θ = p² - 1/p² + 1

Solution

p² - 1/p² + 1 = (sec θ + tan θ)² -1/(sec θ + tan θ)² +1

= sec²θ + tan²θ + 2secθ·tanθ -1/ sec²θ + tan²θ + 2secθ·tanθ +1

= (sec²θ -1) + tan²θ + 2secθ·tanθ /sec²θ + 2tanθ secθ + 1·(1 + tan²θ)

= tan²θ + tan²θ + 2secθ ·tanθ / sec²θ + 2secθ·tanθ + sec²θ

= 2 tan²θ + 2 secθ·tanθ /2sec²θ + 2 secθ·tanθ

= 2 tanθ (tanθ + secθ) / 2sec θ(secθ + tanθ)

= tanθ / secθ

= sinθ / cosθ·sinθ

= sin θ

Hence,

Proved !!