If tanα = 2
find \(\frac{cosec^2α-sec^2α}{cosec^2α+sec^2α}\)+\(\frac{3}{5}\)
Answer & explanation
Correct answer: option 4
tanα = \(\frac{2}{1}\)=\(\frac{P}{B}\)
H = \(\sqrt {(2)^2+(1)^2}\)
H = \(\sqrt {5}\)
⇒ \(\frac{(\frac{\sqrt {5}}{2})^2-(\frac{\sqrt {5}}{1})^2}{(\frac{\sqrt {5}}{2})^2+(\frac{\sqrt {5}}{1})^2}\)+\(\frac{3}{5}\)
⇒ \(\frac{\frac{5}{4}-5}{\frac{5}{4}+5}\)+\(\frac{3}{5}\)
⇒ \(\frac{\frac{-15}{4}}{\frac{25}{4}}\)+\(\frac{3}{5}\) = \(\frac{-3}{5}\) +\(\frac{3}{5}\)
= 0