Prove that \[\sqrt{\dfrac{1+\cos A}{1-\cos A}}=\operatorname{cosec}A+\cot A\]
ANSWER
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- Hint: In this question, let us multiply the numerator and denominator with suitable terms under the square root. Then we obtain square terms by using the trigonometric identities in both numerator and denominator such that the square root gets cancelled. Now, on simplifying further we get the result. Following identities can be used to solve: \[\cot \theta =\dfrac{1}{\tan \theta }\] \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
Complete step-by-step solution -
TRIGONOMETRIC IDENTITIES: An equation involving trigonometric functions which is true for all those angles for which the functions are defined is called trigonometric identity. Some of those identities are \[\cot \theta =\dfrac{1}{\tan \theta }\] \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] Now, given in the question that, \[\Rightarrow \sqrt{\dfrac{1+\cos A}{1-\cos A}}\] Now, let us multiply the numerator and denominator with suitable terms so that it can be simplified further. Let us multiply the numerator and denominator with \[1+\cos A\] then we get, \[\Rightarrow \sqrt{\dfrac{1+\cos A}{1-\cos A}\times \dfrac{1+\cos A}{1+\cos A}}\] As we already know from algebra that, \[\begin{align} & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \\ & a\times a={{a}^{2}} \\ \end{align}\] Now, by using the formulae mentioned the above expression can be rewritten as: \[\Rightarrow \sqrt{\dfrac{{{\left( 1+\cos A \right)}^{2}}}{1-{{\cos }^{2}}A}}\] Now, by using the trigonometric identity the denominator can be expressed as \[\Rightarrow \sqrt{\dfrac{{{\left( 1+\cos A \right)}^{2}}}{{{\sin }^{2}}A}}\text{ }\left[ \because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right]\] Now, this expression can be further written as \[\Rightarrow \sqrt{{{\left( \dfrac{1+\cos A}{\sin A} \right)}^{2}}}\] Let us now cancel the square term and square root term and then further write the expression. \[\Rightarrow \dfrac{1+\cos A}{\sin A}\] Now, by dividing the sine term with each of the term in the numerator it can be further written as \[\Rightarrow \dfrac{1}{\sin A}+\dfrac{\cos A}{\sin A}\] As we already know from the trigonometric identities that \[\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }\] \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\] \[\cot \theta =\dfrac{1}{\tan \theta }\] Now, the above equation can be further written as \[\Rightarrow \operatorname{cosec}A+\cot A\]
Note: Instead of multiplying the numerator and denominator with \[1+\cos A\]we can also multiply it with \[1-\cos A\]which results in interchanging the terms that we got which means sine term in the numerator and cosine term in the denominator. Then as the denominator consists of addition terms we need to further simplify it by multiplying with suitable terms again which in turn becomes a lengthy process. It is important to apply the respective trigonometric identity because neglecting any term or using wrong identity changes the result completely.