Question

(b)
Without using trigonometrical tables, evaluate:
5 cos 0° - 2 sin 30° + V3 cos 30°
- +3 sin 29° sec 61°
tan 30° x tan 60° x cos 60°​

Answer 1

$\dfrac{5\cos 0-2\sin30+\sqrt{3}\cos 30}{\tan30\times \tan60\times \cos60+ 3\sin29 \sec61}=\dfrac{11}{7}$

Step-by-step explanation:

We have,

$\dfrac{5\cos 0-2\sin30+\sqrt{3}\cos 30}{\tan30\times \tan60\times \cos60+ 3\sin29 \sec61}$

To find, $\dfrac{5\cos 0-2\sin30+\sqrt{3}\cos 30}{\tan30\times \tan60\times \cos60+ 3\sin29 \sec61}=?$

∴ $\dfrac{5\cos 0-2\sin30+\sqrt{3}\cos 30}{\tan30\times \tan60\times \cos60+ 3\sin29 \sec61}$

$=\dfrac{5(1)-2(\dfrac{1}{2} )+\sqrt{3}(\dfrac{\sqrt{3}}{2} )}{\dfrac{1}{\sqrt{3}} \times \sqrt{3} \times (\dfrac{1}{2})+ 3\sin29 \sec (90-29)}$

We know that,

The trigonometric identity,

$\sin30= \dfrac{1}{2}$, $\cos30 =\dfrac{\sqrt{3}}{2}$ , $\cos 0=1$, $\cos60=\dfrac{1}{2}$,

$\tan30=\dfrac{1}{\sqrt{3}}$ and $\tan60=\sqrt{3}$

$=\dfrac{5-1+\dfrac{3}{2}}{\dfrac{1}{2}+ 3\sin29 \csc 29}$

Using the trigonometric identity,

$\sec (90-A)=\csc A$

$=\dfrac{4+\dfrac{3}{2}}{\dfrac{1}{2}+ 3\sin29\dfrac{1}{\sin29}}$

$=\dfrac{\dfrac{8+3}{2}}{\dfrac{1}{2}+ 3(1)}$

$=\dfrac{\dfrac{11}{2}}{\dfrac{1+6}{2}}$

$=\dfrac{\dfrac{11}{2}}{\dfrac{7}{2}}$

$=\dfrac{11}{7}$

Hence, $\dfrac{5\cos 0-2\sin30+\sqrt{3}\cos 30}{\tan30\times \tan60\times \cos60+ 3\sin29 \sec61}=\dfrac{11}{7}$