What is the value of $\frac{1-tan^245^o}{1+cos^290^o}$ ?
Answer & explanation
Correct answer: option 1
\(\frac{1 - tan² 45º}{1 + cos²90º}\)
= \(\frac{1 - 1}{1 + 0}\)
= 0
What is the value of $\frac{1-tan^245^o}{1+cos^290^o}$ ?
Correct answer: option 1
\(\frac{1 - tan² 45º}{1 + cos²90º}\)
= \(\frac{1 - 1}{1 + 0}\)
= 0