Question

Given that $$\Theta$$ is an obtuse angle measured in radians and that $$\sin \Theta =k$$,find ,in terms of k ,an expression for

(i)$$\cos \Theta$$

(ii)$$\tan \Theta$$

(iii)$$\sin \left ( \Theta +\pi \right )$$

(i)$$\Theta$$ is obtuse,$$\sin \Theta =k$$

$$\cos \Theta =-\sqrt{\left ( 1-k^{2} \right )}$$

(ii)$$\tan \Theta =\frac{\sin \Theta }{\cos \Theta }$$

$$\rightarrow \tan \Theta =-\frac{k}{\sqrt{\left ( 1-k^{2} \right )}}$$

(iii)$$\sin \left ( \Theta +\pi \right )=-k$$

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 $$\frac{1-sin2x}{1+sin2x}=\frac{1}{3}\rightarrow sin2x=\frac{1}{2} x=\frac{\Pi }{12} x=\frac{5\Pi }{12}$$