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Question: Prove the following expression: \[\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \se...

Prove the following expression:
(cosec⁡θ−sin⁡θ)(sec⁡θ−cos⁡θ)(tan⁡θ+cot⁡θ)=1\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)=1

Explanation

Solution

Hint: First of all, take LHS and convert the whole expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta . Now simplify the whole equation and use identity sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 to get the desired result.

Complete step- by-step-by-step solution -
In this question, we have to prove that
(cosec⁡θ−sin⁡θ)(sec⁡θ−cos⁡θ)(tan⁡θ+cot⁡θ)=1\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)=1
First of all, let us take the LHS of the above equation.
E=(cosec⁡θ−sin⁡θ)(sec⁡θ−cos⁡θ)(tan⁡θ+cot⁡θ)E=\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)
We know that cosec⁡x=1sin⁡x\operatorname{cosec}x=\dfrac{1}{\sin x}. By using this in the above expression, we get,
E=(1sin⁡θ−sin⁡θ)(sec⁡θ−cos⁡θ)(tan⁡θ+cot⁡θ)E=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \sec \theta -\operatorname{cos}\theta \right)\left( \tan \theta +\cot \theta \right)
Also, we know that sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x}. By using this in the above expression, we get,
E=(1sin⁡θ−sin⁡θ)(1cos⁡θ−cos⁡θ)(tan⁡θ+cot⁡θ)E=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \dfrac{1}{\cos \theta }-\cos \theta \right)\left( \tan \theta +\cot \theta \right)
Now, we know that tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x} and cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}. By using this in the above expression, we get,
E=(1sin⁡θ−sin⁡θ)(1cos⁡θ−cos⁡θ)(sin⁡θcos⁡θ+cos⁡θsin⁡θ)E=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \dfrac{1}{\cos \theta }-\cos \theta \right)\left( \dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta } \right)
By simplifying the above expression, we get,
E=(1−sin⁡2θsin⁡θ)(1−cos⁡2θcos⁡θ)(sin⁡2θ+cos⁡2θsin⁡θcos⁡θ)E=\left( \dfrac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta } \right)
We know that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. By using this in the above expression, we get,
E=(1−sin⁡2θsin⁡θ)(1−cos⁡2θcos⁡θ)(1sin⁡θcos⁡θ)E=\left( \dfrac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{1}{\sin \theta \cos \theta } \right)
Also, we know that 1−sin⁡2x=cos⁡2x1-{{\sin }^{2}}x={{\cos }^{2}}x and 1−cos⁡2x=sin⁡2x1-{{\cos }^{2}}x={{\sin }^{2}}x. By using this in the above expression, we get,
E=(cos⁡2θsin⁡θ).(sin⁡2θcos⁡θ)(1sin⁡θcos⁡θ)E=\left( \dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right).\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{1}{\sin \theta \cos \theta } \right)
We can also write the above equation as
E=(cos⁡2θ).(sin⁡2θ)(cos⁡2θ).(sin⁡2θ)E=\dfrac{\left( {{\cos }^{2}}\theta \right).\left( {{\sin }^{2}}\theta \right)}{\left( {{\cos }^{2}}\theta \right).\left( {{\sin }^{2}}\theta \right)}
By canceling the like terms from RHS of the above equation, we get,
E = 1
So, we get LHS = RHS
Hence, we have proved
(cosec⁡θ−sin⁡θ)(sec⁡θ−cos⁡θ)(tan⁡θ+cot⁡θ)=1\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)=1

Note: In these types of questions which have multiple trigonometric ratios, it is always advisable to convert the whole expression in terms of sin⁡θ\sin \theta and cos⁡θ\cos \theta especially when RHS is a constant value like in the above question. Also, we not only need to remember sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1 but its various forms like sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x or cos⁡2x=1−sin⁡2x{{\cos }^{2}}x=1-{{\sin }^{2}}x also. We should not get confused with multiple formulas.