Question

if sec thita-tan thita=p, then obtain values of tan thita ,sec thita, sin thita in terms of p.​

Answer 1

Step-by-step explanation:

I am taking (A) instead of theta

secA - tanA = p-----------eq1

rationalise

(secA - tanA)×(secA + tanA)/(secA + tanA) = p

(sec^2A - tan^2A)/(secA + tanA) = p

{sec^2A =1 + tan^2A}

{sec^2A - tan^2A = 1}

1/(secA + tanA) = p

secA + tanA = 1/p-------eq2

solve equations 1 and 2

secA - tanA = p

secA + tanA = 1/p

-----------------------------

2secA - 0 = p + 1/p

2secA = (p^2 +1)/p

secA = (p^2 +1)/2p

put the value of sec in equation 1 or 2

secA + tanA = 1/p

tanA = secA - 1/p

tanA = (p^2 +1)/2p - 1/p

tanA = (p^2 - 1)/2p

tanA = (p^2 - 1)/2p

sinA/cosA = (p^2 - 1)/2p (tanA = sinA/cosA)

sinA×secA = (p^2 - 1)/2p (secA = 1/cosA)

sinA × (p^2 + 1)/2p = (p^2 - 1)/2p

sinA = (p^2 - 1)/(p^2 + 1)