CAIE A-Level Further Math A2 2.1.3 Hyperbolic Functions QuestionsPractise proving and applying hyperbolic identities from exponential definitions with Further Mathematics Paper 2 questions.Syllabus2028–2030CourseFurther Mathematics 9231LevelA2
Exam pointsProve and apply identities linking hyperbolic functions using their exponential definitions.
CAIE A-Level Further Math A2 2.1.3 Hyperbolic Functions Questions question 1[Maximum number: 3]Starting from the definitions of tanh and sech in terms of exponentials, prove thattanh2t+sech2t=1.\tanh ^{2} t+\operatorname{sech}^{2} t=1 .tanh2t+sech2t=1.Show Answertanht=et−e−tet+e−t\tanh t=\frac{\mathrm{e}^{t}-\mathrm{e}^{-t}}{\mathrm{e}^{t}+\mathrm{e}^{-t}} \quadtanht=et+e−tet−e−t sech t=2et+e−tt=\frac{2}{\mathrm{e}^{t}+\mathrm{e}^{-t}}t=et+e−t2(et−e−tet+e−t)2+4(et+e−t)2=e2t+e−2t+2(et+e−t)2=1\left(\frac{\mathrm{e}^{t}-\mathrm{e}^{-t}}{\mathrm{e}^{t}+\mathrm{e}^{-t}}\right)^{2}+\frac{4}{\left(\mathrm{e}^{t}+\mathrm{e}^{-t}\right)^{2}}=\frac{\mathrm{e}^{2 t}+\mathrm{e}^{-2 t}+2}{\left(\mathrm{e}^{t}+\mathrm{e}^{-t}\right)^{2}}=1(et+e−tet−e−t)2+(et+e−t)24=(et+e−t)2e2t+e−2t+2=1Writes in a single fraction over commondenominator, AG.Withold A1 mark if they used mixedvariables within a single line of working.3Add to Test