Question

The value of cos(60+2A)÷cos 2A-√3 sin 2A

Answer 1

Answer:

$\huge\bold\red{\frac{1}{2}}$

Step-by-step explanation:

For simplicity, let's denote angle A as

$\bold{alpha \: ( \alpha )}$

Now, we have to find the value of,

$\frac{ \cos(60 + 2 \alpha ) }{ \cos(2 \alpha ) - \sqrt{3} \sin(2 \alpha ) }$

But, we know that,

cos (A+B) = cosA cosB - sinA sinB

therefore, using this identity,

we get,

$= \frac{ \cos(60) \cos(2 \alpha ) - \sin(60) \sin(2 \alpha ) }{ \cos(2 \alpha ) - \sqrt{3} \sin(2 \alpha ) }$

Putting the value of,

$\bold{cos60=\frac{1}{2}\:and\:sin60=\frac{\sqrt{3}}{2}}$

we get,

$= \frac{ \frac{ \cos(2 \alpha ) }{2} - \frac{ \sqrt{3} \sin(2 \alpha ) }{2} }{ \cos(2 \alpha ) - \sqrt{3} sin(2 \alpha) } \\ \\ = \frac{1}{2 } \times \frac{ \cos(2 \alpha ) - \sqrt{3} \sin(2 \alpha ) }{ \cos(2 \alpha ) - \sqrt{3} \sin(2 \alpha ) } \\ \\ = \frac{1}{2}$