Question

2+ sin45+ cos45+ sec30- tan 30

Answer 1

Step-by-step explanation:

$theefore \: \sin(45) = \frac{1}{ \sqrt{2} } \\ \cos(45) = \frac{1}{ \sqrt{2} } \\ \sec(30) = \frac{2}{ \sqrt{3} } \\ \tan(30) = \frac{1}{ \sqrt{3} }$

Keep the above values in the sum

$2 + \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } + \frac{2}{ \sqrt{3} } - \frac{1}{ \sqrt{3} }$

$2 + \frac{2}{ \sqrt{2} } + \frac{2}{ \sqrt{3} } - \frac{1}{ \sqrt{3} }$

$2 + \frac{2 \sqrt{3} + 2 \sqrt{2} }{ \sqrt{6} } - \frac{1}{ \sqrt{3} }$

$\frac{2 \sqrt{6 } + 2 \sqrt{3} + 2 \sqrt{2} - \sqrt{2} }{ \sqrt{6} }$

$\frac{2 \sqrt{6} + 2 \sqrt{3} + \sqrt{2} }{ \sqrt{6} }$

The future processe in the above photo

Answer 2

To Find:

2+ sin45+ cos45+ sec30- tan 30

Answer:

values of trigonometric ratios

• sin45° = $\frac{1}{ \sqrt{2} }$

• cos 45° =$\frac{1}{ \sqrt{2} }$

• sec 30°=$\frac{2}{ \sqrt{3} }$

• tan 30°=$\frac{1}{ \sqrt{3} }$

2+ sin45+ cos45+ sec30- tan 30

$\longrightarrow$$2 + \frac{1}{ \sqrt{2} } + \frac{1}{ \sqrt{2} } + \frac{2}{ \sqrt{3} } - \frac{1}{ \sqrt{3} }$

$\implies$$\frac{2 \sqrt{6} + \sqrt{3} + \sqrt{3} + 4 - 2 }{ \sqrt{6} }$

$\implies$$\frac{2 \sqrt{6} + 2 \sqrt{3} + 2 }{ \sqrt{6} }$

hence , ans:

$\huge \tt \underline \orange {\frac{2 \sqrt{6} \:+ \:2 \sqrt{3}\: + \:2 }{ \sqrt{6} }}$