Question

Refer to the image provided above



Evaluate and find the value of question above.

Answer 1

Given,

$\frac{\tan^{2}60\textdegree.4\cos^{2}45\textdegree+3\sec^{2}30\textdegree+5\cos^{2}90}{cosec30\textdegree+\sec60\textdegree-\cot30\textdegree}\\\;\\=\frac{(\sqrt{3})^{2}.4\times(\frac{1}{\sqrt{2}})^{2}+3(\frac{2}{\sqrt{3}})^{2}+5\times0}{(2)+(2)-(\sqrt{3})^{2}}\\\;\\=\frac{3\times2\;+4+0}{2+2-3}\\\;\\=\frac{6+4}{4-3}\\\;\\=\frac{10}{1}\\\;\\=10\;\;\textbf{Ans.}$

Answer 2

sin theta = height / hypotenuse

cos theta = base / hypotenuse

tan theta = height /base

cot theta = base / height

sec theta = hypotenuse / base

cosec theta = hypotenuse / height.

To evaluate ,

(tan^2 60°.4cos^2 45° + 3sec^2 30° + 5cos^2 90°)/cosec 30° + sec 60° - cot^2 30°

We know ,

tan 60° = √3

cos 45° = 1/√2

sec 30° = 2/√3

cos 90° = 0

cosec 30° = 2

sec 60° = 2

cot 30° = √3

These all values are taken from the trigonometric table.

By putting these values in the given expression , we get the answer as :-

10

Pls refer to the attachment for detailed solution.