Question

Root 1-sin thita/1+sin thita is equal to sec thita - sin thita

Answer 1

To prove :-

√(1 - sinθ /1 + sinθ) = secθ - sinθ.

Proof :-

$\underline{\underline{\sf{LHS:-}}}$

$\displaystyle{\tt{ \sqrt{\frac{1-sin \theta}{1+sin \theta} } }}$

$\displaystyle{\tt{ \to \sqrt{\frac{(1-sin \theta)(1-sin \theta)}{(1+sin \theta)(1-sin \theta)} } }}\\\\\displaystyle{\tt{ \to \sqrt{\frac{(1-sin\theta)^2}{(1)^2-(sin\theta)^2} } }}\\\\\displaystyle{\tt{ \to \sqrt{\frac{(1-sin\theta)^2}{1-sin^2\theta} } }}\\\\\displaystyle{\tt{ \to \sqrt{\frac{(1-sin\theta)^2}{cos^2\theta} } }}\\\\\displaystyle{\tt{ \to \sqrt{\bigg( \frac{1-sin\theta}{cos\theta}\bigg)^2 } }}\\\\\displaystyle{\tt{ \to \frac{1-sin\theta}{cos\theta} }}$

$\displaystyle{\tt{ \to \frac{1}{cos\theta}-\frac{sin\theta}{cos\theta} }}\\\\\displaystyle{\tt{ \to sec\theta+tan\theta=}} \underline{\underline{\sf{RHS}}}$

Answer 2

Step-by-step explanation:

according to me there must be tan theta instead of sin theta in RHS

if wrong , sorry