Question

if 4 x is equal to cos theta and 4 upon x is equal to cos theta then find the value of 4 in bracket x square minus one upon x square ​

Answer 1

if 4 x is equal to cos theta and 4 upon x is equal to cos theta then find the value of 4 in bracket x square minus one upon x square ....

I think your question is ---> if 4x = cosθ and 4/x = cosθ, then find the value of 4(x² - 1/x²)

solution : it is given that,

4x = cosθ ⇒x = cosθ/4

and 4/x = cosθ ⇒1/x = cosθ/4

so, 4(x² - 1/x²)

= 4[(cosθ/4)² - (cosθ/4)² ]

= 4 [cos²θ/16 - cos²θ/16 ]

= 4/16 [ cos²θ - cos²θ ]

= 1/4 × 0

= 0

hence, the value 4(x² - 1/x²) = 0

Answer 2

Answer:

$4(x^{2}-\frac{1}{x^{2}})=0$

Step-by-step explanation:

In the question,

We have,

cosθ = 4x

also,

cosθ = 4/x

So,

We need to find the value of,

$4(x^{2}-\frac{1}{x^{2}})......(1)$

Therefore,

From the given data we can say that,

4x = 4/x

So,

$\frac{4}{x}=4x\\So,\\4x^{2}=4\\x^{2}=1\\x=-1,1$

Now, on putting the value of 'x' in the equation (1) we get,

$4(x^{2}-\frac{1}{x^{2}})=4((1)^{2}-\frac{1}{1^{2}})=4(1-1)=0$

Therefore, on finally putting the value of x as -1 or 1, the value of the equation is given by,

$4(x^{2}-\frac{1}{x^{2}})=0$