Question

Without using tables evaluate : -
5 cos o' - 2 sin 30° + V 3 cos 30'/
Tan 30° X tan 60° X cos 60°
+
3 sin 29 sec 61
... ​

Answer 1

Answer:

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Answer 2

$\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}$