Question

# 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

Hint: Use the information that, $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and ${\sin ^2}\theta + {\cos ^2}\theta = 1$.

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.