Question

If secA + tanA = p, prove that secA=1/2(p+1/p)​

Answer 1

Answer :

Proved!

Step-by-step explanation :

Given that :

SecA + tanA = p.

To Prove :

SecA = 1/2(p+1/p)​.

Solution :

$\sf As\;given,\\ \\\implies secA+tanA=p......Eq(1)\\ \\ \\\textbf{\underline{\underline{We know that:-}}}\\ \\ \\\bigstar\;\boxed{\sf sec^2A-tan^2A=1}}\\ \\So,\\ \\\implies (secA+tanA)(secA-tanA)=1\\ \\\implies p(secA-tanA)=1\\ \\\implies secA-tanA=\dfrac{1}{p}$

$\textbf{\underline{\underline{Add both Equations,}}}\\ \\ \sf\implies 2secA=p+\dfrac{1}{p}\\ \\\implies secA=\dfrac{(p^2+1)}{2p}\\ \\\implies SecA=\dfrac{1}{secA}=\dfrac{2p}{(p^2+1)}\\ \\ \\\implies \sqrt{\dfrac{1-4p^2}{(p^2+1)^2}}\\ \\ \\\implies \sqrt{\dfrac{(p^2+1)^2- 4p^2}{(p^2+1)^2}}\\ \\ \\\implies \sqrt{\dfrac{(p^4+2p^2+1-4p^2)}{(p^2+1)}}\\ \\ \\\implies \sqrt{\dfrac{(p^2-1)^2}{(p^2+1)}}\\ \\ \\\implies \dfrac{(p^2-1)}{(p^2+1)}\\ \\ \\\\\bigstar\textbf{\underline{\underline{Hence, Proved}}}$