1.

Sintheta.cosec theta/costheta=​

Answer»

rove :

We NEED to PROVE that, \dfrac{1-\sin\theta}{1+\sin\theta}=(\sec\theta-\tan\theta)^2

Solution,

Taking LHS, we GET :

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

Rationalizing the denominator we get,

\begin{gathered}\dfrac{1-\sin\theta}{1+\sin\theta}\times \dfrac{1-\sin\theta}{1-\sin\theta}\\\\=\dfrac{(1-\sin\theta)^2}{1-\sin^2\theta}\\\\=\dfrac{(1-\sin\theta)^2}{\cos^2\theta}\ (\because 1-\sin^2\theta=\cos^2\theta)\\\\\\=(\dfrac{1-\sin\theta}{\cos\theta})^2\\=(\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{\cos\theta})^2\\\\=(\sec\theta-\tan\theta)^2\\\\=RHS\end{gathered}

So,

LHS = RHS

Hence, proved.


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