Short Question 9 & 10

Question 9:

Solution:
(a)


$\sin \theta =\frac{3}{5}\cos \theta =\frac{4}{5}\tan \theta =\frac{3}{4}$


(c)
$\begin{array}{l}\tan \left({360}^{\circ }+\theta \right)\\ \\ =\frac{\tan {360}^{\circ }+\tan \theta }{1-\tan {360}^{\circ }\tan \theta }\\ \\ =\frac{0+\tan \theta }{1-\left(0\right)\left(\tan \theta \right)}\\ =\tan \theta \\ =\frac{3}{4}\end{array}$

$\begin{array}{l}\text{LHS:}{\cot }^{2}x-{\cot }^{2}x{\cos }^{2}x\\ ={\cot }^{2}x\left(1-{\cos }^{2}x\right)\\ ={\cot }^{2}x\left(si{n}^{2}x\right)\\ =\frac{{\cos }^{2}x}{si{n}^{2}x}\left(si{n}^{2}x\right)\\ ={\cos }^{2}x\text{(RHS)}\end{array}$


(b)
$\begin{array}{l}\text{LHS:}\frac{\sec x}{\sec x-\cos x}\\ =\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-\cos x}=\frac{\frac{1}{\cos x}}{\frac{1}{\cos x}-\frac{{\cos }^{2}x}{\cos x}}\\ =\frac{\frac{1}{\cos x}}{\frac{1-{\cos }^{2}x}{\cos x}}=\frac{1}{\cos x}\times \frac{\cos x}{1-{\cos }^{2}x}\\ =\frac{1}{1-{\cos }^{2}x}=\frac{1}{si{n}^{2}x}\\ =\cos e{c}^{2}x\text{(RHS)}\end{array}$