Question 31


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Jambmaths question: 

If $\tan \theta =\tfrac{4}{3}$, calculate ${{\sin }^{2}}\theta -{{\cos }^{2}}\theta $

Option A: 


Option B: 


Option C: 


Option D: 


Jamb Maths Solution: 

$\begin{align}  & \text{From the definition of tangent of an angle} \\ & \tan \theta =\frac{opposite}{adjacent}=\frac{4}{3} \\ & A{{C}^{2}}={{3}^{3}}+{{4}^{2}}\text{    (Pythagoras theorem)} \\ & AC=5 \\ & \cos \theta =\tfrac{adj}{hyp}=\tfrac{3}{5},\text{   }{{\cos }^{2}}\theta ={{(\tfrac{3}{5})}^{2}}=\tfrac{9}{25} \\ & \sin \theta =\tfrac{opp}{hyp}=\tfrac{4}{5},\text{    }{{\sin }^{2}}\theta ={{(\tfrac{4}{5})}^{2}}=\tfrac{16}{25} \\ & {{\sin }^{2}}\theta -{{\cos }^{2}}\theta =\tfrac{16}{25}-\tfrac{9}{25}=\tfrac{7}{25} \\\end{align}$

Jamb Maths Topic: 
Year of Exam: 


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