Evaluate the following: $$\dfrac{5cos^260° + 4sec^230° – tan^245°}{sin^2 30° + cos^2 30°}$$

Asked by Pragya Singh | 1 year ago |  76

##### Solution :-

We know that,

cos 60° = $$\dfrac{1}{2}$$

sec 30° = $$\dfrac{2}{\sqrt{3}}$$

tan 45° = 1

sin 30° = $$\dfrac{1}{2}$$

cos 30° = $$\dfrac{\sqrt{3}}{2}$$

Now, substitute the values in the given problem, we get

$$\dfrac{(5cos^260° + 4sec^230° – tan^245°)}{(sin^2 30° + cos^2 30°)}$$

= 5 $$(\dfrac{1}{2})^2$$+4 $$( \dfrac{2}{\sqrt{3}})^2-$$$$\dfrac{-1^2}{(\dfrac{1}{2^2})}$$+($$\dfrac{\sqrt{3}}{2}$$)2

$$\dfrac{5}{4}+\dfrac{16}{3-1}$$

$$\dfrac{1}{4}+\dfrac{3}{4}$$

$$\dfrac{15+64-12}{12}$$$$\dfrac{4}{4}$$

$$\dfrac{67}{12}$$

Answered by Abhisek | 1 year ago

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