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I tried it a lot but am not able to get this.Pls help in how should I think when solving this type of question and which side is better to try to simplify first (LHS or RHS).Please share the solution in that way.

$$ \frac{1-\sin A}{1+\sin A} = 1 + 2 \tan A \left(\tan A - \sec A \right) $$

One of the ways I tried but am not understand that how to simplify it in such a way that you get the RHS?

2 Answers2

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Starting from the LHS we have $$\frac{1-\sin(A)}{1+\sin(A)}$$ $$=\frac{1-\sin(A)}{1+\sin(A)}\cdot\frac{1-\sin(A)}{1-\sin(A)}=\frac{1-2\sin(A)+\sin^2(A)}{1-\sin^2(A)}$$ $$=\frac{\cos^2(A)-2\sin(A)+2\sin^2(A)}{\cos^2(A)}$$

using the identity $\sin^2(A)+\cos^2(A)=1$. Can you end it now?

Alessio K
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  • Yes thank you sir –  Oct 12 '20 at 08:00
  • What strategy did you think of before solving it? –  Oct 12 '20 at 08:02
  • You’re welcome! First to rationalise the denominator, then since the denominator simplifies to $\cos^2(x)$ and we have a $1$ on the RHS, I applied the identity to get the $\cos^2(x)$ in the numerator so we could cancel (which also introduced another $2$ in the numerator). – Alessio K Oct 12 '20 at 08:11
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You have $LHS$ in terms of sine. So it's better to prove that $RHS = LHS$ as $\tan$ and $\sec$ can be converted into terms of $\cos$ and $ \sin$.
$\begin{align}1+2\tan A(\tan A - \sec A) &= 1+2\dfrac{\sin A(\sin A- 1)}{\cos^2A} \\&= \dfrac{\cos^2A+2\sin A(\sin A- 1)}{\cos^2A} \\&=\dfrac{(\cos^2A+\sin^2A)+\sin^2A-2\sin A}{1-\sin^2A} \\&=\dfrac{(1-\sin A)^2}{1-\sin^2A} = \color{blue}{\dfrac{1-\sin A}{1+\sin A}}\end{align}$

(Provided $A \ne (2n+1)\frac\pi2, n\in \Bbb Z$)

19aksh
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  • What strategy did you think of before solving it? –  Oct 12 '20 at 08:01
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    First of all $\tan$ and $\sec$ in product provides $\cos^2 A$ in the denominator. Which can be written as $1-\sin^2A = (1-\sin A)(1+\sin A).$ This indicated, I need $(1-\sin A)^2$ in the numerator so as to get the desired result. – 19aksh Oct 12 '20 at 08:05