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Solve the following trigonometric identity.
$1 + {\cot ^2}\theta = $
A. ${\sec ^2}\theta $
B. ${\text{cose}}{{\text{c}}^2}\theta $
C. ${\tan ^2}\theta $
D. 1

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Last updated date: 19th Jul 2024
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Answer
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Hint: Use the information that, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

Complete step-by-step answer:

We have given $1 + {\cot ^2}\theta $. We can write $\cot \theta $ as $\dfrac{{\cos \theta }}{{\sin \theta }}$. So, let’s substitute the values and solve,
$
  1 + {\cot ^2}\theta \\
   = 1 + \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} \\
   = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }} \\
   = \dfrac{1}{{{{\sin }^2}\theta }} \\
   = {\text{cose}}{{\text{c}}^2}\theta \\
$
Hence, the required value of $1 + {\cot ^2}\theta $ is ${\text{cose}}{{\text{c}}^2}\theta $.

Note: Such problems could be easily solved by altering the basic trigonometric identities. Mistakes can be avoided in conversion of one trigonometric function to another.