Question

(i)Prove the identity $$\frac{\sin \Theta }{1-\cos \Theta }-\frac{1}{\sin \Theta }=\frac{1}{\tan \Theta }$$.

(ii) Hence solve the equation $$\frac{\sin \Theta }{1-\cos \Theta }-\frac{1}{\sin \Theta }=4\tan \Theta$$ for $$0^{\circ}< \Theta < 180^{\circ}$$.

(i)$$LHS=\frac{\sin ^{2}\Theta -(1-\cos \Theta )}{\left ( 1-\cos \Theta \right )\sin \Theta }$$\

$$=\frac{1-\cos ^{2}\Theta-1+\cos \Theta }{\left ( 1-\cos \Theta \right )\sin \Theta }$$

$$=\frac{\cos \Theta \left ( 1-\cos \Theta \right ) }{\left ( 1-\cos \Theta \right )\sin \Theta }$$

$$=\frac{1}{\tan \Theta }$$

(ii)$$\tan \Theta =\pm \frac{1}{2}$$

$$26.6^{\circ},153.4^{\circ}$$

Question

(a)The diagram shows part of the graph of $$y=a+b\sin x$$.Find the values of constant a and b.

(b)(i)Show that the equation $$\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta -\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )$$

may be expressed as $$3\cos ^{2}\Theta -2\cos \Theta -1=0$$.

(c)Hence solve the equation $$\left ( \sin \Theta +2\cos \Theta \right )\left ( 1+\sin \Theta-\cos \Theta \right )=\sin \Theta \left ( 1+\cos \Theta \right )$$ for $$-180^{\circ}\leq \Theta \leq 180^{\circ}$$

(i)a=-2,b=3

(b)(ii)$$s+s^{2}-sc+2c+2sc=s+sc\rightarrow s^{2}-2c^{2}+2c=0$$

$$1-\cos ^{2}\Theta -2\cos ^{2}\Theta +2\cos \Theta =0$$

$$3\cos ^{2}\Theta -2\cos \Theta -1=0$$

(ii)$$\cos \Theta = 1$$or $$-\frac{1}{3}$$

$$\Theta =0^{\circ}$$ or $$109.5^{\circ}$$ or $$-109.5^{\circ}$$

Question

(i) Show that the equation $$\frac{4\cos \Theta }{\tan \Theta }+15=0$$ can be expressed as $$4\sin ^{2}\Theta -15\sin \Theta -4=0$$.

(ii) Hence solve the equation $$\frac{4\cos \Theta }{\tan \Theta }+15=0$$ for $$0^{\circ}\leq \Theta \leq 360^{\circ}$$.

(i)$$4\cos ^{2}\Theta +15\sin \Theta =0$$

$$4(1-s^{2})+15s=0\rightarrow 4\sin ^{2}\Theta -15\sin \Theta -4=0$$

(ii)$$\sin \Theta =-\frac{1}{4}$$

$$\Theta$$=194.5 or 345.5

#### Question

Solve the equation $$3 sin^2\Theta = 4 cos \Theta -1$$ for 0° < 0 ≤ 360°.

$$3\sin ^{2}\Theta =4\cos \Theta -1$$

Uses $$s^{2}+c^{2}=1$$

$$\rightarrow 3c^{2}+4c-4\left ( =0 \right )$$

$$\left ( \rightarrow c=\frac{2}{3} or -2\right )$$

$$\rightarrow \Theta =48.2^{\circ}$$ or $$311.8^{\circ}$$

$$\left ( 0.268\pi ,1.73\pi \right )$$

Question

(i) Prove the identity $$\left ( \frac{1}{cos}-tanx \right )^{2}\equiv \frac{1-sinx}{1+sinx}$$

(ii) Hence solve the equation $$\left ( \frac{1}{cosx}-tan2x \right )^{2}=\frac{1}{3}$$ for0 ≤ x ≤ 0.

(i) LHS=$$\left ( \frac{1}{c} -\frac{s}{c}\right )^{2}=\frac{(1-s)(1-s)}{c^{2}}$$
=$$\frac{(1-s)(1-s)}{1-s^{2}}$$
=$$\frac{(1-s)(1-s)}{(1-s)(1+s)}$$
$$\frac{1-sinx}{1+sinx}$$

(ii) Uses part (i) to obtain $$\frac{1-sin2x}{1+sin2x}=\frac{1}{3}\rightarrow sin2x=\frac{1}{2} x=\frac{\Pi }{12} x=\frac{5\Pi }{12}$$

### Question

(a) Show that $$\frac{\sin \theta }{1-\sin \theta }-\frac{\sin \theta }{1+\sin \theta }=2\tan ^{2}\theta$$  [3]

(b) Hence solve the equation $$\frac{\sin \theta }{1-\sin \theta }-\frac{\sin \theta }{1+\sin \theta }=8, for\ 0^{o}< \theta < 180^{o}$$  [3]

Ans

7 (a) $$\left ( \frac{\sin \theta }{1-\sin \theta } -\frac{\sin \theta }{1+\sin \theta }\right )=\frac{\sin \theta (1+\sin \theta )-\sin \theta (1-\sin \theta )}{1-\sin ^{2}\theta }$$

$$\frac{2\sin ^{2}\theta }{\cos ^{2}\theta }$$

2 tan2 θ

7 (b) $$2\tan ^{2}\theta =8\rightarrow \tan \theta =(\pm )2$$

( θ= ) =63.4°, 116.6°

### Question

Solve the equation
$$\frac{tan \theta + 3 sin \theta + 2}{tan \theta – 3 sin \theta + 1$$
for $$0^o ≤ \theta ≤ 90^o$$.

Ans:

$$2tan \theta – 6 sin \theta + 2 = tan \theta + 3 sin \theta + 2 \rightarrow tan \theta – 9 sin \theta (=0)$$
$$sin \theta – 9 sin \theta cos \theta (=0)$$
$$sin \theta (1-9 cos \theta )(=0) \rightarrow sin \theta =0, cos \theta = \frac{1}{9}$$
$$\theta = 0$$ or $$83.6^o$$ (only answers in the given range)

### Question.

(a).  Prove the indentity  $$\frac{1+sin\Theta }{cos\Theta }+\frac{cos\Theta }{1+sin\Theta}\equiv \frac{2}{cos\Theta }$$

(b) Hence solve the equation  $$\frac{1+sin\Theta }{cos\Theta }+\frac{cos\Theta }{1+sin\Theta}\equiv \frac{2}{cos\Theta }$$  for $$0 \leq \Theta \leq 2\Pi$$.

(a)  $$\frac{(1+\sin \theta )^{2}+\cos ^{2}\theta }{\cos \theta (1+\sin \theta )}$$
Use of $$sin2 θ + cos2 θ = 1 \rightarrow \frac{2+2\sin \theta }{\cos \theta (1+\sin \theta )}\rightarrow \frac{2}{\cos \theta }$$
(b)  $$\frac{2}{\cos \theta }=\frac{3}{\sin \theta }\rightarrow \tan \theta =1.5$$