If $ \cos \theta = \frac{a}{b} $, find $ 1 + \tan^2 \theta $

Assessment: JAMB UTME - Mathematics - 1986 Subject: General Mathematics

Question 1 Report

\frac{b^{2}}{a^{2}}

\frac{a^{2}}{b^{2}}

\frac{a^{2} + b^{2}}{b^{2} - a^{2}}

\frac{2 a^{2} + b^{2}}{a^{2} + b^{2}}

Answer Details

cos $θ$ = $\frac{a}{b}$ , Sin $θ$ = $\sqrt{\frac{b^{2} - a^{2}}{a}}$
Tan $θ$ = $\sqrt{\frac{b^{2} - a^{2}}{a^{2}}}$ , Tan 2 = $\sqrt{\frac{b^{2} - a^{2}}{a^{2}}}$
1 + tan2 $θ$ = 1 + $\frac{b^{2} - a^{2}}{a^{2}}$
= $\frac{a^{2} + b^{2} - a^{2}}{a^{2}}$
= $\frac{b^{2}}{a^{2}}$

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If \( \cos \theta = \frac{a}{b} \), find \( 1 + \tan^2 \theta \)

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If \( \cos \theta = \frac{a}{b} \), find \( 1 + \tan^2 \theta \)

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