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Question: Prove the following identities, where the angles involved are acute angles for which the expressions...

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(csc⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ{\left( {\csc \theta - \cot \theta } \right)^2} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}

Explanation

Solution

To prove the identity, we simplify the L.H.S of the equation using trigonometric ratios, we will convert the cosec and cot to the basic trigonometric terms i.e. sine and cosine as we have a lot of formulas for sine and cosine. Further we will use the algebraic identities to simplify the terms.

Complete step-by-step answer:
Given data, angles involved are acute angles.
To prove: (csc⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ{\left( {\csc \theta - \cot \theta } \right)^2} = \dfrac{{1 - \cos \theta }}{{1 + \cos \theta }}
We start off by solving the L.H.S of the equation,
Using the trigonometric ratios, we know
csc⁡θ=1sin⁡θ\csc \theta = \dfrac{1}{{\sin \theta }}
And
cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }}
Let us now substitute these values in the LHS of the given equation to proceed forward and prove the equation the same as RHS.
=(csc⁡θ−cot⁡θ)2 =(1sin⁡θ−1tan⁡θ)2  = {\left( {\csc \theta - \cot \theta } \right)^2} \\\ = {\left( {\dfrac{1}{{\sin \theta }} - \dfrac{1}{{\tan \theta }}} \right)^2} \\\
As we know the value of tangent of angle in terms of sine and cosine is given by the formula:
tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}
Let us now substitute this formula in the above equation,
=(1sin⁡θ−cos⁡θsin⁡θ)2= {\left( {\dfrac{1}{{\sin \theta }} - \dfrac{{\cos \theta }}{{\sin \theta }}} \right)^2}
Let us further solve the equation by taking LCM in the LHS.
=(1−cos⁡θsin⁡θ)2 =(1−cos⁡θ)2sin⁡2θ  = {\left( {\dfrac{{1 - \cos \theta }}{{\sin \theta }}} \right)^2} \\\ = \dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{{\sin }^2}\theta }} \\\
Using the trigonometric identity relating the square of sine of the angle to the cosine of the angle we will proceed further.
=(1−cos⁡θ)21−cos⁡2θ[∵sin⁡2θ+cos⁡2θ=1]= \dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{1 - {{\cos }^2}\theta }}\left[ {\because {{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right]
Now let us use the algebraic formula to solve the given term by expanding the denominator,
We know that:
a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)
Comparing this to the denominator we have:

=(1−cos⁡θ)212−cos⁡2θ =(1−cos⁡θ)2(1+cos⁡θ)(1−cos⁡θ)[∵a2−b2=(a+b)(a−b)]  = \dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{{1^2} - {{\cos }^2}\theta }} \\\ = \dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}}\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \\\

Now, let us cancel the common term in numerator and the denominator to get the final answer.

=(1−cos⁡θ)2(1+cos⁡θ)(1−cos⁡θ) =(1−cos⁡θ)(1+cos⁡θ) =RHS  = \dfrac{{{{\left( {1 - \cos \theta } \right)}^2}}}{{\left( {1 + \cos \theta } \right)\left( {1 - \cos \theta } \right)}} \\\ = \dfrac{{\left( {1 - \cos \theta } \right)}}{{\left( {1 + \cos \theta } \right)}} \\\ = RHS \\\

Hence proved

Note: The key in solving such types of problems is to convert the given trigonometric functions into different functions, i.e. cosec to sine and cot to tangent. On converting it is easy to simplify the equation using the trigonometric identities and formulae. In order to solve such types of problems, students must remember the basic trigonometric formulas.