Question

tan theta – cot theta =a
and sintheta+cos theta = b.
then (b^2-1)^2(a^2+4) is equal to​


I will mark u as brainliest if you answer it

Answer 1

Answer:

4

Step-by-step explanation:

Given that

$tan\theta- cot\theta = a$

and $sin\theta + cos\theta = b$

Thus

$a^{2}= (tan\theta-cot\theta)^{2}$

or, $a^{2} = (tan^{2}\theta+cot^{2}\theta-2tan\theta.cot\theta )$

or, $a^{2} = (tan^{2}\theta+cot^{2}\theta-2 )$

Therefore $a^{2} +4=tan^{2}\theta+cot^{2}\theta+2$

or $a^{2} +4=(1+tan^{2} \theta)+(1+cot^{2} \theta)$

or, $a^{2} +4 = sec^{2}\theta+ cosec^{2}\theta$

or, $a^{2} +4 = \frac{1}{cos^{2}\theta } + \frac{1}{sin^{2}\theta}$

or, $a^{2} +4 = \frac{sin^{2}\theta + cos^{2}\theta}{sin^{2}\theta.cos^{2}\theta }$

or, $a^{2} +4 = \frac{1}{sin^{2}\theta.cos^{2}\theta}$

Now, $b^{2} = (sin\theta+cos\theta)^{2}$

or, $b^{2} = (sin^{2}\theta + cos^{2}\theta + 2sin\theta.cos\theta)$

or, $b^{2} = 1+2sin\theta.cos\theta$

Therefore, $b^{2} -1 = 2sin\theta.cos\theta$

And $(b^{2}-1)^{2} = 4sin^{2} \theta.cos^{2} \theta$

Now, $(b^{2}-1)^{2} (a^{2} +4) = 4sin^{2} \theta.cos^{2} \theta.\frac{1}{sin^{2} \theta.cos^{2}\theta}$

or, $(b^{2}-1)^{2} (a^{2} +4) = 4$