How do you simplify the expression #(sec^2t-1)/tant#? Trigonometry Trigonometric Identities and Equations Fundamental Identities 1 Answer Rajat Chatterjee Nov 3, 2016 ans.#tant#. Explanation: #(sec^2t-1)/tant#=#tan^2t/tant#=#tant#. Answer link Related questions How do you use the fundamental trigonometric identities to determine the simplified form of the... How do you apply the fundamental identities to values of #theta# and show that they are true? How do you use the fundamental identities to prove other identities? What are even and odd functions? Is sine, cosine, tangent functions odd or even? How do you simplify #sec xcos (frac{\pi}{2} - x )#? If #csc z = \frac{17}{8}# and #cos z= - \frac{15}{17}#, then how do you find #cot z#? How do you simplify #\frac{\sin^4 \theta - \cos^4 \theta}{\sin^2 \theta - \cos^2 \theta} # using... How do you prove that tangent is an odd function? How do you prove that #sec(pi/3)tan(pi/3)=2sqrt(3)#? See all questions in Fundamental Identities Impact of this question 2783 views around the world You can reuse this answer Creative Commons License