Question

If sec theta - tan theta=√3-2,
show that 1-sin theta/1+sin theta=7-4√3.Pls.give me crct answer ☺

Answer 1

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Answer 2

Answer and Explanation:

Given : $\sec \theta - \tan \theta=\sqrt{3}-2$

To show : $\frac{1-\sin\theta}{1+\sin\theta}=7-4\sqrt3$

Solution :

Taking LHS,

$LHS=\frac{1-\sin\theta}{1+\sin\theta}$

Multiply and divide by $1-\sin\theta$

$=\frac{1-\sin\theta}{1+\sin\theta}\times \frac{1-\sin\theta}{1-\sin\theta}$

$=\frac{(1-\sin\theta)^2}{1^2-\sin^2\theta}$

$=\frac{1-2\sin\theta+sin^2\theta}{\cos^2\theta}$

$=\frac{1}{\cos^2\theta}-\frac{2\sin\theta}{\cos^2\theta}+\frac{sin^2\theta}{\cos^2\theta}$

$=sec^2\theta-2\sec\theta\tan^2\theta+tan^2\theta$

$=(sec\theta-tan\theta)^2$

We know,  $\sec \theta - \tan \theta=\sqrt{3}-2$ substitute

$=(\sqrt{3}-2)^2$

$=\sqrt{3}^2+2^2-2\times 2\times\sqrt3$

$=3+4-4\sqrt3$

$=7-4\sqrt3$

=RHS

Hence proved.