Question

Prove that. sinA +1 - cosA/sinA -1+cosA = 1+ sinA / cosA

Answer 1

I hope that you understand all steps.....

Answer 2

Answer:  The proof is done below.

Step-by-step explanation:  We are given to prove the following trigonometric equality :

$\dfrac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}=\dfrac{1+\sin\theta}{\cos\theta}.$

We will be using the following formulas :

$(i)~\dfrac{\sin\theta}{\cos\theta}=\tan\theta,\\\\\\(ii)~\sec\theta=\dfrac{1}{\cos\theta},\\\\\\(iii)~\sec^2\theta-\tan^2\theta=1\\\\(iii)~a^2-b^2=(a+b)(a-b).$

The proof is as follows :

$\dfrac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}\\\\\\=\dfrac{\dfrac{\sin\theta+1-\cos\theta}{\cos\theta}}{\dfrac{\sin\theta-1+\cos\theta}{\cos\theta}}\\\\\\=\dfrac{\tan\theta+\sec\theta-1}{\tan\theta-\sec\theta+1}\\\\\\=\dfrac{\tan\theta+\sec\theta-(\sec^2\theta-\tan^2\theta)}{\tan\theta-\sec\theta+1}\\\\\\=\dfrac{\tan\theta+\sec\theta-(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{\tan\theta-\sec\theta+1}\\\\\\=\dfrac{(\tan\theta+\sec\theta)(1-\sec\theta+\tan\theta)}{1-\sec\theta+\tan\theta}\\\\\\=\tan\theta+\sec\theta\\\\\\=\dfrac{sin\theta}{\cos\theta}+\dfrac{1}{\cos\theta}\\\\\\=\dfrac{1+\sin\theta}{\cos\theta}\\\\=R.H.S.$

Thus, we get

$\dfrac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}=\dfrac{1+\sin\theta}{\cos\theta}.$

Hence proved.