 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/9/solution

sin θ = 2 t / (1 + t^{2})

and

cos θ = (1 - t^{2}) / (1 + t^{2})

. Dividing and then taking the reciprocal square of the cosine gives

tan θ = \frac{2 t}{1 - t ^{2}}, sec^{2} θ = \frac{( 1 + t ^{2} ) ^{2}}{( 1 - t ^{2} ) ^{2}} .

These formulas apply where

t

is defined and

t \neq = \pm 1

; at

t = \pm 1

the original tangent and secant function are undefined because

cos θ = 0

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