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Question: How do you prove \[{\csc ^2}\theta - {\cot ^2}\theta = \cot \theta \tan \theta \]...

How do you prove csc⁡2θ−cot⁡2θ=cot⁡θtan⁡θ{\csc ^2}\theta - {\cot ^2}\theta = \cot \theta \tan \theta

Explanation

Solution

To prove the given trigonometric function, apply trigonometric identity formulas i.e., applying reciprocal trigonometric functions of cosec, cot and tan functions, 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 function is
⇒csc⁡2θ−cot⁡2θ=cot⁡θtan⁡θ\Rightarrow {\csc ^2}\theta - {\cot ^2}\theta = \cot \theta \tan \theta
To prove the function let us use the following reciprocal identities
⇒csc⁡θ=1sin⁡θ\Rightarrow \csc \theta = \dfrac{1}{{\sin \theta }} …………….. 1
⇒cot⁡θ=cos⁡θsin⁡θ\Rightarrow \cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }} ……………… 2
⇒tan⁡θ=sin⁡θcos⁡θ\Rightarrow \tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} ……………… 3
Substitute the identities of equation 1, 2 and 3 with respect to the function asked in the given equation as
⇒csc⁡2θ−cot⁡2θ=cot⁡θtan⁡θ\Rightarrow {\csc ^2}\theta - {\cot ^2}\theta = \cot \theta \tan \theta
⇒1sin⁡2θ−cos⁡2θsin⁡2θ=cos⁡θsin⁡θ⋅sin⁡θcos⁡θ\Rightarrow \dfrac{1}{{{{\sin }^2}\theta }} - \dfrac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }} = \dfrac{{\cos \theta }}{{\sin \theta }} \cdot \dfrac{{\sin \theta }}{{\cos \theta }}
Simplifying the functions, we get
⇒1−cos⁡2θsin⁡2θ=1\Rightarrow \dfrac{{1 - {{\cos }^2}\theta }}{{{{\sin }^2}\theta }} = 1
We know that
⇒1−cos⁡2θ\Rightarrow 1 - {\cos ^2}\theta = sin⁡2θ{\sin ^2}\theta
Hence, applying this formula we get
⇒sin⁡2θsin⁡2θ=1\Rightarrow \dfrac{{{{\sin }^2}\theta }}{{{{\sin }^2}\theta }} = 1
Which implies
⇒1=1\Rightarrow 1 = 1

Therefore, LHS = RHS i.e., csc⁡2θ−cot⁡2θ=cot⁡θtan⁡θ{\csc ^2}\theta - {\cot ^2}\theta = \cot \theta \tan \theta

Additional Information:
The three basic functions in trigonometry are sine, cosine and tangent. Based on these three functions the other three functions that are cotangent, secant and cosecant are derived.
All the trigonometrical concepts are based on these functions. Hence, to understand trigonometry further we need to learn these functions and their respective formulas at first.
If θ is the angle in a right-angled triangle, then
Sin θ = perpendicularhypotenuse\dfrac{{perpendicular}}{{hypotenuse}}
Cos θ = basehypotenuse\dfrac{{base}}{{hypotenuse}}
Tan θ = perpendicularbase\dfrac{{perpendicular}}{{base}}
The other three functions i.e., cot, sec and cosec depend on tan, cos and sin respectively.

Note: The key point to prove any trigonometric function is to note the formulas of all related functions with respect to the equation, and prove all the terms by considering LHS terms and prove for RHS and to prove the given function we must know the reciprocal formula of cosec and hence, by applying this and simplifying the terms of the function we can prove LHS = RHS.