Question

prove the identity sin(2x)/sinx-cos(2x)/cosx= secx​

Answer 1

The identity $\frac{sin(2x)}{sin\ x} -\frac{cos(2x)}{cos\ x} = sec\ x$ is proved below.

What are trigonometric functions?

Trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths.

We have,

$\frac{sin(2x)}{sin\ x} -\frac{cos(2x)}{cos\ x} = sec\ x$

Now,

Take Left hand side;

i.e.

$=\frac{sin(2x)}{sin\ x} -\frac{cos(2x)}{cos\ x}$

Now,

Using $Sin(2x)= 2Sin\ x \ Cos\ x$

And,

$Cos(2x)=Cos^2x-Sin^2x$

So,

We get,

$=\frac{2Sin\ x \ Cos\ x}{sin\ x} -\frac{(Cos^2x-Sin^2x)}{cos\ x}$

Now, Simplify,

$=\ 2 Cos\ x -\frac{(Cos^2x-Sin^2x)}{cos\ x}$

We get,

$=\ \frac{2 Cos^2 x-Cos^2x+Sin^2x}{cos\ x}$

Simplify,

$=\ \frac{Cos^2 x+Sin^2x}{cos\ x}$

Now, Using the identity,

$Cos^2 x+Sin^2x=1$

We get,

$=\ \frac{1}{cos\ x}=Sec\ x$

That means Left hand side is equals to Right hand side.

i.e. $Sec\ x=Sec\ x$

Hence we can say that the identity $\frac{sin(2x)}{sin\ x} -\frac{cos(2x)}{cos\ x} = sec\ x$ is proved above.

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