Question

Prove that \[1 + {\cot ^2}\theta = {\csc ^2}\theta \]

Answer 1

Hint: Here we need to prove that \[1 + {\cot ^2}\theta = {\csc ^2}\theta \], in which \[1 + {\cot ^2}\theta \] can be written in terms of sin and cosec, as we know that sec, csc and cot are derived from primary functions of sin, cos and tan, hence using these functions we can prove that LHS = RHS of the given function.

Complete step-by-step solution:
The given equation is
\[1 + {\cot ^2}\theta = {\csc ^2}\theta \]
In which we need to prove that LHS terms are equal to RHS terms as given.
Let us consider the LHS terms of the given equation i.e.,
\[1 + {\cot ^2}\theta \]
Here, \[1 + {\cot ^2}\theta \] can be written in terms of sin and cosec as:
\[1 + {\cot ^2}\theta = 1 + \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\]
As we know that \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\], hence applying this in the equation becomes as
\[1 + {\cot ^2}\theta = \dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta }}\]
We know that \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], hence simplifying the terms we get
\[1 + {\cot ^2}\theta = \dfrac{1}{{{{\sin }^2}\theta }}\]
Since, \[\dfrac{1}{{\sin \theta = \csc\theta }}\] we get
\[1 + {\cot ^2}\theta = {\csc ^2}\theta \] = RHS

Hence proved LHS = RHS

Additional information:
In trigonometry sin, cos and tan values are the primary functions we consider while solving trigonometric problems. These trigonometry values are used to measure the angles and sides of a right-angle triangle. Apart from sine, cosine and tangent values, other values are cotangent, secant and cosecant. The trigonometric values are about the knowledge of standard angles for a given triangle as per the trigonometric ratios. Trigonometric ratios are Sine, Cosine, Tangent, Cotangent, Secant and Cosecant.

Note: The key point to prove any trigonometric function is that we must note the chart of all functions as shown and calculate all the terms asked. And here are some of the identities to be noted while solving the trigonometric function.
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\], \[\dfrac{1}{{{{\sin }^2}\theta }} = {\csc ^2}\theta \], \[\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}\], \[\sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }}\], \[\csc\theta = \dfrac{{\sec \theta }}{{\tan \theta }}\]