JAMB - Mathematics (1986 - No. 30)
If cos\(\theta\) = \(\frac{a}{b}\), find 1 + tan2\(\theta\)
\(\frac{b^2}{a^2}\)
\(\frac{a^2}{b^2}\)
\(\frac{a^2 + b^2}{b^2 - a^2}\)
\(\frac{2a^2 + b^2}{a^2 + b^2}\)
توضیح
cos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\)
Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\)
1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\)
= \(\frac{a^2 + b^2 - a^2}{a^2}\)
= \(\frac{b^2}{a^2}\)
Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\)
1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\)
= \(\frac{a^2 + b^2 - a^2}{a^2}\)
= \(\frac{b^2}{a^2}\)