Question

Find the value of (cosec² θ – l).tan²θ
How many zeros does the polynomial (x – 3)2 – 4 can have?​

Answer 1

Hey user. The answer to your first question is 1.

From the trigonometric identity 1+cot^2 theta=cosec^2theta we get,

cosec^2-1=cot^2theta

We know that cot^2theta=1/tan^2theta.

Substituting cot^2theta in the given equation

(cosec^2theta-1)*tan^2theta=cot^2theta*tan^2theta=tan^2theta/tan^2theta=1.

For your second question the answer is 2.

This is because the given equation is a quadratic equation which can at most have two zeroes.

Let's check it out.

(x-3)^2-4=

x^2+9-6x-4=x^2+5-6x

By splitting the middle term (-6x) we get

x^2-x-5x+5

x(x-1)-5(x-1)

This means (x-5) and (x-1) are its factors.

equating them to 0 we get to know that

x=5,1

Therefore the answer to your second question is 5 and 1 are the zeroes if the given quadratic polynomial

Hope it was helpful to you.

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