Question

1/1 sin theta 1/1-sin theta=2sec2 theta

Answer 1

$\textbf{To prove:}$

$\dfrac{1}{1+sin\theta}+\dfrac{1}{1-sin\theta}=2\,sec^2\theta$

$\textbf{Solution:}$

$\text{Consider,}$

$\dfrac{1}{1+sin\theta}+\dfrac{1}{1-sin\theta}$

$=\dfrac{1-sin\theta+1+sin\theta}{(1+sin\theta)(1-sin\theta)}$

$\text{Using,}\;\boxed{\bf(a+b)(a-b)=a^2-b^2}$

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

$\text{Using,}\;\boxed{\bf\,1-sin^2A=cos^2A}$

$=\dfrac{2}{cos^2\theta}$

$=2(\dfrac{1}{cos^2\theta})$

$=2\,sec^2\theta$

$\therefore\bf\dfrac{1}{1+sin\theta}+\dfrac{1}{1-sin\theta}=2\,sec^2\theta$

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