sec squared 55 - tan squared 55

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