1+\(^{ }\tan^{2^{ }}\alpha\)= \(1+\left(\dfrac{sin\alpha}{cos\alpha}\right)^2\)=\(\dfrac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}\)=\(\dfrac{1}{cos^2\alpha}\)
b: \(1+tan^2a=\dfrac{1}{cos^2a}=1+\dfrac{9}{25}=\dfrac{34}{25}\)
=>cos^2a=25/34
=>\(cosa=\dfrac{5}{\sqrt{34}}\)
\(sina=\sqrt{1-\dfrac{25}{34}}=\dfrac{3}{\sqrt{34}}\)