Question

if (sec theta - 1)(sec theta + 1) = 1/3 then cos theta =?

Answer 1

       

$(\sec\theta-1)(\sec\theta+1)=\frac{1}{3} \\ \\ \\ \Rightarrow\ \sec^2\theta-1=\frac{1}{3}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{(a-b)(a+b)=a^2-b^2} \\ \\ \\ \Rightarrow\ \sec^2\theta=\frac{1}{3}+1 \\ \\ \\ \Rightarrow\ \sec^2\theta=\frac{4}{3}\\ \\ \\ \Rightarrow\ \frac{1}{\sec^2\theta}=\frac{1}{\frac{4}{3}} \\ \\ \\ \Rightarrow\ \cos^2\theta=\frac{3}{4} \\ \\ \\ \Rightarrow\ \cos\theta=\sqrt{\frac{3}{4}} \\ \\ \\ \Rightarrow\ \cos\theta=\ \ \bold{\pm\frac{\sqrt{3}}{2}}$

$Here, \\ \\ \\ \theta=30(n+6),\ \ for any integer \ \ n.$

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