tan θ=P-14p⇒sec θ-tan θ= (1) 2por12p (2) 12por-2p - Maths -

Question

tan   θ = P - 1 4 p ⇒ sec   θ - tan
  θ =
(1) 2 p or 1 2 p (2) 1 2 p or - 2 p (3) - 1 2 p or   2 p (4) -
1 2 p or   - 2 p

Neha Sethi · Accepted Answer

Dear student
Given: tanθ=p-14pConsider,secθ-tanθ=1+tan2θ-tanθ    As sec2A=1+tan2A=1+p-14p2-p+14p=1+p2+116p2-12-p+14p=p2+116p2+12-p+14p=p+14p2-p+14p=±p+14p-p+14p=p+14p-p+14p or -p-14p-p+14p=12p or -2p
Regards

tan θ = P - 1 4 p ⇒ sec θ - tan θ = (1) 2 p or 1 2 p (2) 1 2 p or - 2 p (3) - 1 2 p or 2 p (4) - 1 2 p or - 2 p

$\tan θ = P - \frac{1}{4 p} \Rightarrow \sec θ - \tan θ =$
(1) $2 p (or) \frac{1}{2 p}$ (2) $\frac{1}{2 p} (or) - 2 p$ (3) $- \frac{1}{2 p} (or) 2 p$ (4) $- \frac{1}{2 p} (or) - 2 p$