Prove the following trigonometric expression:

${(1 + {cot ^2}A)}{sin ^2}A = 1$

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Hint: It would be better to start proving this from the LHS since it is more complex than RHS. Let’s apply a basic trigonometric identity and we’ll prove it in only a few steps. 

Complete step by step answer:

We have the left hand side term as,

$(1 + {\cot ^2}A){\sin ^2}A{\text{ }} --{\text{Eq 1}}$

Now, we know that,

$\cos e{c^2}A - {\cot ^2}A = 1$

${\cot ^2}A + 1 = \cos e{c^2}A{\text{ }} --{\text{Eq 2}}$

Putting equation (2) in equation (1),we get,

$\cos e{c^2}A{\sin ^2}{\text{A}}$

$ =\dfrac{1}{{{{\sin }^2}{\text{A}}}}{\sin ^2}{\text{A}}$

$= {\text{1 }}$


Which is the required right hand side term.

Therefore, LHS = RHS

Hence proved

Note: To prove a trigonometric identity, we always start from either the (LHS) or (RHS) and apply the identities step by step until we reach the other side. However, it's always better to start from the more complex side as it is a lot easier to eliminate terms to make a complex function simpler than to find ways to introduce terms stuck in between.