[3 marks sum]

Question 513 Marks

Given that $\tan \left(\theta_1+\theta_2\right)$ Given that $\tan \left(\theta_1+\theta_2\right)$ $=\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}$ Find $(\theta _1 + \theta _2)$ when $\tan \theta _1 = \frac{1}{2}$ and $\tan \theta_2=\frac{1}{3}$.

Answer

We have,
$\tan \theta _1 = =\frac{1}{2}$ and $\tan \theta_2=\frac{1}{3}$
$\therefore \tan \left(\theta_1+\theta_2\right)=\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \tan \theta_2}$
$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}$
$=\frac{\frac{5}{6}}{1-\frac{1}{6}}$
$=\tan \left(\theta_1+\theta_2\right)=\frac{\frac{5}{6}}{\frac{5}{6}}$
$=\tan \left(\theta_1+\theta_2\right)=1$
$=\theta_1+\theta_2=45^{\circ}$

View full question & answer→

Question 523 Marks

Solve the following equation:
$\frac{\cos ^2 \theta-3 \cos \theta+2}{\sin ^2 \theta}=1$.

Answer

We have,
$\frac{\cos ^2 \theta-3 \cos \theta+2}{\sin ^2 \theta}=1$
$\Rightarrow \cos^2 \theta - 3 \cos \theta + 2 = \sin^2 \theta$
$\Rightarrow \cos^2 \theta - 3 \cos \theta + 2 - \sin^2 \theta = 0$
$\Rightarrow \cos^2\theta - 3 \cos \theta + 1 + \cos^2 \theta = 0$
$\Rightarrow 2cos^2\theta - 3 \cos \theta + 1 = 0$
$\Rightarrow 2cos^2\theta - 2 \cos \theta - \cos \theta + 1 = 0$
$\Rightarrow 2cos \theta ( \cos \theta - 1) - 1( \cos \theta - 1) = 0$
$\Rightarrow (\cos \theta - 1)(2 \cos \theta - 1) = 0$
$\Rightarrow \cos \theta - 1 = 0$ or $\Rightarrow 2 \cos \theta - 1 = 0$
$\Rightarrow \cos \theta = 1$ or $\Rightarrow \cos \theta = \frac{1}{2}$
$\Rightarrow \theta = 0^\circ$ or $\Rightarrow \theta = 60^\circ$
$Since, 0 < \theta < 90^\circ$
So, $\theta = 60^\circ$ is the solution of the equation.

View full question & answer→

Question 533 Marks

Solve the following equation:
$\frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4$

Answer

We have,
$\therefore \frac{\cos \theta}{1-\sin \theta}+\frac{\cos \theta}{1+\sin \theta}=4$
$\Rightarrow \cos \theta\left\{\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}\right\}=4$
$\Rightarrow \cos \theta\left\{\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}\right\}=4$
$\Rightarrow 2 \cos \theta = 4( 1 - \sin \theta )( 1 + \sin \theta )$
$\Rightarrow 2 \cos \theta = 4( 1 - \sin^2\theta )$
$\Rightarrow 2 \cos \theta = 4cos^2\theta$
$\Rightarrow 4 \cos^2\theta - 2 \cos \theta = 0$
$\Rightarrow 2 \cos \theta ( 2 \cos \theta - 1) = 0$
$\Rightarrow 2 \cos \theta = 0$ or $\Rightarrow 2 \cos \theta - 1 = 0$
$\Rightarrow \cos \theta=0 \text { or } \Rightarrow \cos \theta=\frac{1}{2}$
$\Rightarrow \theta=60^{\circ},\left(\text { since } 0<\theta<90^{\circ}\right) .$

View full question & answer→

Mathematics [3 marks sum]