Question

How do you verify
\[\]${{\sec }^{2}}\left( x \right)+{{\csc }^{2}}\left( x \right)=\left[ \sec^2 \left( x \right)\csc^2 {{\left( x \right)}} \right]?$

Answer 1

Hint: The function such as sine, cosine and tangent of an angle refer as a basic trigonometric function. And on other side second $\left( \sec \right)$ constant $\left( \csc \right)$ and cotangent $\left( \cot \right)$ are functions which is reciprocal of cosine. Sine and tangent respectively in the given problem convert are all the functions such as $\sec ,\csc ,\cot $and $\tan $ to $\sin $ and$\cos $, most of this can be done ring quotient and reciprocal identities. After that check all the angles for sums and differences. So, you have to use proper identity for removing them. Also check for angle multiple and then use the appropriate formula for removing it.

Complete step by step solution:
Let’s start with LHS,
\[{{\sec }^{2}}\left( x \right)+{{\sec }^{2}}\left( x \right)\]
\[=\dfrac{1}{{{\cos }^{2}}\left( x \right)}+\dfrac{1}{{{\sin }^{2}}\left( x \right)}\]

\[=\dfrac{{{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)}{{{\cos }^{2}}\left( x \right).{{\sin }^{2}}\left( x \right)}\]
\[=\dfrac{1}{{{\cos }^{2}}\left( x \right).\sin^2 \left( x \right)}\]
\[=\dfrac{1}{{{\cos }^{2}}\left( x \right)}.\dfrac{1}{\sin^2 \left( x \right)}\]
\[={{\sec }^{2}}\left( x \right).\csc^2 \left( x \right)\]
\[=\left[ \sec^2 \left( x \right)\csc^2 \left( x \right) \right]\]
Hence,
The given equation \[{{\sec }^{2}}\left( x \right)+{{\csc }^{2}}\left( x \right)=\left( \sec^2 \left( x \right)\csc ^2{{\left( x \right)}} \right)\] is verified.

Note: You have to get both sides of the equation in the same function. You don’t have to always use $\sin $ and $\cos $, but it is easier to complete when both sides are composed of similar functions. Make sure that your entire angle must be the same. And also same for addition and substation. Use the identity. If you want to remove the power use the${{\cos }^{2}}\left( x \right)+{{\sin }^{2}}\left( x \right)=1$.