Question 9

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Jambmaths question: 

If $\tan \theta =\tfrac{5}{4}$, find ${{\sin }^{2}}\theta -{{\cos }^{2}}\theta$

Option A: 

$\tfrac{41}{9}$

Option B: 

$\tfrac{9}{41}$

Option C: 

1

Option D: 

$\tfrac{5}{4}$

Jamb Maths Solution: 

$\begin{align}  & A{{C}^{2}}={{5}^{2}}+{{4}^{2}} \\ & A{{C}^{2}}=41 \\ & AC=\sqrt{41} \\ & \sin \theta =\frac{5}{\sqrt{41}} \\ & \cos \theta =\frac{4}{\sqrt{41}} \\ & {{\sin }^{2}}\theta -{{\cos }^{2}}\theta ={{\left( \frac{5}{\sqrt{41}} \right)}^{2}}-{{\left( \frac{4}{\sqrt{41}} \right)}^{2}} \\ & {{\sin }^{2}}\theta -{{\cos }^{2}}\theta =\frac{25}{41}-\frac{16}{41}=\frac{9}{41} \\\end{align}$

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