Question

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Answer this question ⤵⤵


$\sf{If \: sin \theta \: + \: \sqrt{sin \theta + \sqrt{sin \theta + \sqrt{sin \theta \: + ........ \infty } } } = {sec}^{4} \alpha },$
$\sf {then \: sin\theta \: is \: equal \: to}$


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First and best = Brainliest

Answer 1

let √sintheta+√sintheta+√sinteta+
x=√sintheta+x
on squaring both sides
xsquare=sintheta+x
xsquare-x-sintheta
use quadratic equation you will get answer as
:-sintheta,1-sintheta
please mark me as brainliest

Answer 2

Answer:

$\boxed{\sf{tan^2\alpha sec^2\alpha}}$

Step-by-step explanation:

Let $\sqrt{sin\theta+\sqrt{sin\theta+\sqrt{sin\theta+....\infty}}}}$ be x .

Then we can write :

$\sqrt{sin\theta+x}=x$

$\implies sin\theta+x=x^2$ ........( 1 )

Also :

$sin\theta+x=sec^4\alpha$

$\implies sec^4\alpha=x^2$ [ from ( 1 ) ]

$\implies sec^2\alpha=x$

Put this in 1

$\implies sin\theta+sec^2\alpha=sec^4\alpha$

$\implies sin\theta=sec^4\alpha-sec^2\alpha$

$\implies sin\theta=sec^2\alpha(sec^2\alpha-1)$

Note that $sec^2\alpha-1=tan^2\alpha$

$\implies sin\theta=sec^2\alpha tan^2\alpha$