If cos\(\theta\) = \(\frac{a}{b}\), find 1 + tan2\(\theta\)
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Correct Answer: Option A
Explanation:
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 + \(\frac{b^2 - a^2}{a^2}\)
= \(\frac{a^2 + b^2 - a^2}{a^2}\)
= \(\frac{b^2}{a^2}\)
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 + \(\frac{b^2 - a^2}{a^2}\)
= \(\frac{a^2 + b^2 - a^2}{a^2}\)
= \(\frac{b^2}{a^2}\)