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Question: Prove that: \[{{\left( \csc \theta -\cot \theta \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta...

Prove that:
(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

Hint: Take the LHS of the expression. Substitute basic trigonometric formulas and convert the expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta . Thus simplify it using basic formulas of sine and cosine function.
Complete step-by-step answer:
We have been given the expression, (csc⁡θ−cot⁡θ)2{{\left( \csc \theta -\cot \theta \right)}^{2}}, which we need to prove is equal to (1−cos⁡θ1+cos⁡θ)\left( \dfrac{1-\cos \theta }{1+\cos \theta } \right).
We know the basic trigonometric formulas,
csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } and cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta }
Now let us substitute these basic trigonometric formulas in the expression, (csc⁡θ−cot⁡θ)2{{\left( \csc \theta -\cot \theta \right)}^{2}}.
(csc⁡θ−cot⁡θ)2=(1sin⁡θ−cos⁡θsin⁡θ)2{{\left( \csc \theta -\cot \theta \right)}^{2}}={{\left( \dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta } \right)}^{2}}
=(1−cos⁡θsin⁡θ)2=(1−cos⁡θ)2(sin⁡θ)2 =(1−cos⁡θ)2sin⁡2θ \begin{aligned} & ={{\left( \dfrac{1-\cos \theta }{\sin \theta } \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta \right)}^{2}}}{{{\left( \sin \theta \right)}^{2}}} \\\ & =\dfrac{{{\left( 1-\cos \theta \right)}^{2}}}{{{\sin }^{2}}\theta } \\\ \end{aligned}
We know that, cos⁡2θ+sin⁡2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1.
Thus we can write that, sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta .
Now let us put the value of sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta in the above expression.
∴(csc⁡θ−cot⁡θ)2=(1−cos⁡θ)21−cos⁡2θ\therefore {{\left( \csc \theta -\cot \theta \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta \right)}^{2}}}{1-{{\cos }^{2}}\theta }
We know the formula, (a−b)(a+b)=a2−b2\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}.
Similarly, 1−cos⁡2θ=12−cos⁡2θ1-{{\cos }^{2}}\theta ={{1}^{2}}-{{\cos }^{2}}\theta
=(1−cos⁡θ)(1+cos⁡θ)=\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)
∴(csc⁡θ−cot⁡θ)2=(1−cos⁡θ)212−cos⁡2θ\therefore {{\left( \csc \theta -\cot \theta \right)}^{2}}=\dfrac{{{\left( 1-\cos \theta \right)}^{2}}}{{{1}^{2}}-{{\cos }^{2}}\theta }
=(1−cos⁡θ)(1−cos⁡θ)(1−cos⁡θ)(1+cos⁡θ)=\dfrac{\left( 1-\cos \theta \right)\left( 1-\cos \theta \right)}{\left( 1-\cos \theta \right)\left( 1+\cos \theta \right)}
Cancel out (1−cos⁡θ)\left( 1-\cos \theta \right) from the numerator and the denominator.
(csc⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ{{\left( \csc \theta -\cot \theta \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }
Thus we proved that, (csc⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ{{\left( \csc \theta -\cot \theta \right)}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }.
Hence proved

Note: To solve this question or the similar type of questions we need to know the basic trigonometric properties and formulas of the functions which are used. If we know these formulas it is easy to solve. Trigonometry is an important portion in mathematics, so learn the basic formulas.