Answer:
sec squared 55 – tan squared 55 = 1
Explanation:
Given, sec square 55 – tan squared 55
We know that,
[tex]\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}[/tex]
And,
[tex]\tan \theta=\frac{\text { perpendicular }}{\text { base }}[/tex]
where Ө is the angle
Substituting the values
[tex]\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}[/tex]
Solving,
[tex]\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}[/tex]
According to Pythagoras theorem,
[tex]\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}[/tex]
Putting this in the equation;
squared 55 - tan squared 55 =
[tex]\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1[/tex]
Therefore, sec squared 55 – tan squared 55 = 1
