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Prove that ${\tan ^2}A + {\cot ^2}A = {\sec ^2} A.{cosec^2}A - 2$

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Hint- Use the trigonometric identities.

We have to prove that ${\tan ^2}A + {\cot ^2}A = {\sec ^2}A.{cosec^2}A - 2$
Now let’s consider the RHS side
We have ${\sec ^2}A.{cosec^2}A - 2$
Now using the trigonometric identity that $\left( {1 + {{\tan }^2}\theta } \right) = {\sec ^2}\theta $and $\left( {1 + {{\cot }^2}\theta = {cosec^2}\theta } \right)$
We can change the RHS side as
$ \Rightarrow \left( {1 + {{\tan }^2}A} \right)\left( {1 + {{\cot }^2}A} \right) - 2$
Let’s simplify this more we get
$1 + {\tan ^2}A + {\cot ^2}A + {\tan ^2}A{\cot ^2}A - 2$
Now ${\cot ^2}A = \dfrac{1}{{{{\tan }^2}A}}$ using this the above gets simplified to
$1 + {\tan ^2}A + {\cot ^2}A + 1 - 2$
$ \Rightarrow {\tan ^2}A + {\cot ^2}A$
Clearly LHS is equal to RHS hence proved

Note- While solving such trigonometric identities problems, we need to have a good grasp over the trigonometric identities, some of them have been mentioned above. It’s always advised to remember them.
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