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# Find the value of $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)$(a) $0$(b) $1$(c) $2$(d) None of these

Last updated date: 15th Sep 2024
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Hint: Use the properties of trigonometric functions and simplify the terms given in the bracket by multiplying each one of them and cancelling out the like terms with opposite signs.

We have to find the value of $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)$. We will begin by simplifying the given expression by multiplying each of the terms in the two brackets.
Thus, we have $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1\left( 1+\cot \theta -\csc \theta \right)+\tan \theta \left( 1+\cot \theta -\csc \theta \right)+\sec \theta \left( 1+\cot \theta -\csc \theta \right)$
Further simplifying the equation, we get $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1+\cot \theta -\csc \theta +\tan \theta \cot \theta +\tan \theta -\tan \theta \csc \theta +\sec \theta +\sec \theta \cot \theta -\sec \theta \csc \theta$
We know that $\tan \theta \cot \theta =1$ and $\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }$.
Thus, we have $\tan \theta \csc \theta =\tan \theta \dfrac{1}{\sin \theta }=\dfrac{1}{\cos \theta }$.
Also, we have $\sec \theta \cot \theta =\cot \theta \dfrac{1}{\cos \theta }=\dfrac{1}{\sin \theta }$.
Similarly, we get $\sec \theta \csc \theta =\dfrac{1}{\sin \theta }\times \dfrac{1}{\cos \theta }=\dfrac{1}{\sin \theta \cos \theta }$.
Substituting all the above equations in the expansion of the given expression, we have $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=1+\cot \theta -\dfrac{1}{\sin \theta }+1+\tan \theta -\dfrac{1}{\cos \theta }+\dfrac{1}{\cos \theta }+\dfrac{1}{\sin \theta }-\dfrac{1}{\sin \theta \cos \theta }$
Thus, we have $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\cot \theta +\tan \theta -\dfrac{1}{\sin \theta \cos \theta }$.
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta }$.
Thus, we have $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\cot \theta +\tan \theta -\dfrac{1}{\sin \theta \cos \theta }=2+\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta \cos \theta }$
Further simplifying the expression, we get $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta }-\dfrac{1}{\sin \theta \cos \theta }$.
We know the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
Thus, we have $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2+\dfrac{1}{\sin \theta \cos \theta }-\dfrac{1}{\sin \theta \cos \theta }$.
So, we get $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)=2$.
Hence, the value of the expression $\left( 1+\tan \theta +\sec \theta \right)\left( 1+\cot \theta -\csc \theta \right)$ is $2$, which is option (c).

Note: Trigonometric functions are real functions which relate any angle of a right angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine and tangent. However, we can also use their reciprocals, i.e., cosecant, secant and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius $1$). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane. One should be careful while using the trigonometric identities and rearranging the terms to convert from one trigonometric function to the other one.