P5.5Pythagorean identities

• M1 for using \(\sec^2\theta = 1 + \tan^2\theta\)
• A1 for \(\sec\theta = \frac{25}{24}\)

Worked solution: \(\sec^2\theta = 1 + \tan^2\theta\) gives \(\sec^2\theta = 1 + \frac{49}{576} = \frac{625}{576}\), so \(\sec\theta = \pm\frac{25}{24}\). As \(\theta\) is acute, \(\sec\theta = \frac{25}{24}\).

• M1 for using \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\)
• A1 for \(\operatorname{cosec}\theta = \frac{25}{7}\)

Worked solution: \(\cot\theta = \frac{24}{7}\), so \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\) gives \(\operatorname{cosec}^2\theta = 1 + \frac{576}{49} = \frac{625}{49}\), so \(\operatorname{cosec}\theta = \pm\frac{25}{7}\). As \(\theta\) is acute, \(\operatorname{cosec}\theta = \frac{25}{7}\).

• M1 for using \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\)
• A1 for \(|\operatorname{cosec}\theta| = \frac{25}{24}\)
• B1 for the correct sign: \(\operatorname{cosec}\theta\) is negative in the third quadrant (\(\sin\theta \lt 0\))

Worked solution: \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\) gives \(\operatorname{cosec}^2\theta = 1 + \frac{49}{576} = \frac{625}{576}\), so \(\operatorname{cosec}\theta = \pm\frac{25}{24}\). \(\operatorname{cosec}\theta\) is negative in the third quadrant (\(\sin\theta \lt 0\)), so \(\operatorname{cosec}\theta = -\frac{25}{24}\).

• M1 for using \(\sec^2\theta = 1 + \tan^2\theta\) (with \(\tan\theta = \frac{24}{7}\))
• A1 for \(|\sec\theta| = \frac{25}{7}\)
• B1 for the correct sign: \(\sec\theta\) is negative in the third quadrant (\(\cos\theta \lt 0\))

Worked solution: \(\tan\theta = \frac{24}{7}\), so \(\sec^2\theta = 1 + \tan^2\theta\) gives \(\sec^2\theta = 1 + \frac{576}{49} = \frac{625}{49}\), so \(\sec\theta = \pm\frac{25}{7}\). \(\sec\theta\) is negative in the third quadrant (\(\cos\theta \lt 0\)), so \(\sec\theta = -\frac{25}{7}\).

• M1 for using \(\operatorname{cosec}^2 2x = 1 + \cot^2 2x\)
• A1* for \(6\cot^2 2x + 5\cot 2x - 6 = 0\) with no errors

Worked solution: \(6(1 + \cot^2 2x) + 5\cot 2x - 12 = 0 \Rightarrow 6\cot^2 2x + 5\cot 2x - 6 = 0\)

• M1 for solving the quadratic: \(\cot 2x = \frac{2}{3}\) or \(-\frac{3}{2}\)
• M1 for \(\tan 2x = \frac{3}{2}\) or \(-\frac{2}{3}\) and finding a principal value for \(2x\)
• A1 for one correct value of \(x\)
• dM1 for a correct method to find further values of \(2x\) in \(0 \le 2x \lt 2\pi\) (adding multiples of \(\pi\))
• A1 for at least 3 correct values of \(x\)
• A1 for all 4 values \(0.491, \ 1.277, \ 2.062, \ 2.848\) and no others

Worked solution: \((3\cot 2x - 2)(2\cot 2x + 3) = 0\) so \(\tan 2x = \frac{3}{2}\) or \(\tan 2x = -\frac{2}{3}\).
For \(0 \le 2x \lt 2\pi\): \(2x = 0.9827\ldots, \ 2.5535\ldots, \ 4.1243\ldots, \ 5.6951\ldots\)
so \(x = 0.491, \ 1.277, \ 2.062, \ 2.848\).