Question

(1-sin theta )(1+sintheta)=​

Answer 1

To prove :

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.

Answer 2

Answer:

cos^2 theta

Step-by-step explanation:

(1 - sin theta)(1 + sin theta)

1^2 - sin^2 theta. [(a+b)(a-b)=a^2 - b^2]

cos^2 theta [sin^2 theta + cos^2 theta= 1

cos^2 theta= 1 - sin^2 theta. ]