Show then solve: algebraic fraction manipulation

A question is this type if and only if part (a) requires manipulating an equation involving algebraic fractions of trig functions (e.g. (tanx+cosx)/(tanx−cosx)=k or 1/(sinθ+cosθ)+1/(sinθ−cosθ)=1) into a standard quadratic trig form, then part (b) solves it.

3 questions · Standard +0.1

1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=1 1.05o Trigonometric equations: solve in given intervals

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CAIE P1 2021 November Q7

8 marks Standard +0.3

Show that the equation $\frac { \tan x + \cos x } { \tan x - \cos x } = k$, where $k$ is a constant, can be expressed as $$( k + 1 ) \sin ^ { 2 } x + ( k - 1 ) \sin x - ( k + 1 ) = 0$$
Hence solve the equation $\frac { \tan x + \cos x } { \tan x - \cos x } = 4$ for $0 ^ { \circ } \leqslant x \leqslant 360 ^ { \circ }$.

CAIE P1 2022 November Q6

6 marks Standard +0.3

Show that the equation $$\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1$$ may be expressed in the form $a \sin ^ { 2 } \theta + b \sin \theta + c = 0$, where $a$, $b$ and $c$ are constants to be found.
Hence solve the equation $\frac { 1 } { \sin \theta + \cos \theta } + \frac { 1 } { \sin \theta - \cos \theta } = 1$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.

CAIE P1 2017 November Q5

7 marks Moderate -0.3

Show that the equation $\frac { \cos \theta + 4 } { \sin \theta + 1 } + 5 \sin \theta - 5 = 0$ may be expressed as $5 \cos ^ { 2 } \theta - \cos \theta - 4 = 0$.
[0pt] [3]
Hence solve the equation $\frac { \cos \theta + 4 } { \sin \theta + 1 } + 5 \sin \theta - 5 = 0$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.