The value of $\frac{sin4θ}{(1-cos4θ)}$ is :
Answer & explanation
Correct answer: option 2
\(\frac{sin4θ}{1 - cos4θ}\)
Put θ = 15º
= \(\frac{sin60º}{1 - cos60º}\)
= \(\frac{ √3/2}{1 - 1/2}\)
= \(\frac{ √3/2}{ 1/2}\)
= √3
The value of $\frac{sin4θ}{(1-cos4θ)}$ is :
Correct answer: option 2
\(\frac{sin4θ}{1 - cos4θ}\)
Put θ = 15º
= \(\frac{sin60º}{1 - cos60º}\)
= \(\frac{ √3/2}{1 - 1/2}\)
= \(\frac{ √3/2}{ 1/2}\)
= √3