Question

If \[p = \left[ {2\sin \theta /\left( {1 + \cos \theta + \sin \theta } \right)} \right]\] and \[q = \left[ {\cos \theta \left( {1 + \sin \theta } \right)} \right]\], then
A. \[pq = 1\]
B. \[q/p = 1\]
C. \[q - p = 1\]
D. \[q + p = 1\]

Answer 1

Hint: In order to Simplify the values of \[p\] and \[q\] separately. Here the values are in fraction, thus we can use rationalization. After simplification check with the options listed, which one is correct. Since there are trigonometric functions, remember the value: \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].

Complete step by step answer:
Given, \[p = \dfrac{{2\sin \theta }}{{1 + \cos \theta + \sin \theta }}\] and \[q = \dfrac{{\cos \theta }}{{1 + \sin \theta }}\].
Let simplify \[p\],
\[p = \dfrac{{2\sin \theta }}{{1 + \cos \theta + \sin \theta }}\]
Here in denominator, numeral and trigonometric functions are present to rationalize, let introduce a bracket to separate numeral and trigonometric functions.
\[p = \dfrac{{2\sin \theta }}{{1 + (\cos \theta + \sin \theta )}}\]
Now rationalize,
\[p= \dfrac{{2\sin \theta }}{{1 + (\cos \theta + \sin \theta )}} \times \dfrac{{1 - (\cos \theta + \sin \theta )}}{{1 - (\cos \theta + \sin \theta )}}\]
\[\Rightarrow p= \dfrac{{2\sin \theta \left( {1 - (\cos \theta + \sin \theta )} \right)}}{{\left( {1 + (\cos \theta + \sin \theta )} \right)\left( {1 - (\cos \theta + \sin \theta )} \right)}}\]
The denominator is in the form \[(a + b)(a - b)\], by applying the formula: \[(a + b)(a - b) = {a^2} - {b^2}\],
\[\Rightarrow p= \dfrac{{2\sin \theta \left( {1 - (\cos \theta + \sin \theta )} \right)}}{{{1^2} - {{(\cos \theta + \sin \theta )}^2}}}\]

Now the denominator is in the form \[{(a + b)^2}\], by applying the formula: \[{(a + b)^2} = {a^2} + {b^2} + 2ab\],
\[p = \dfrac{{2\sin \theta \left( {1 - (\cos \theta + \sin \theta )} \right)}}{{{1^2} - ({{\cos }^2}\theta + {{\sin }^2}\theta + 2\cos \theta \sin \theta )}}\]
We know that \[{1^2} = 1\] and \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] by substituting the values,
\[ p= \dfrac{{2\sin \theta \left( {1 - (\cos \theta + \sin \theta )} \right)}}{{1 - (1 + 2\cos \theta \sin \theta )}}\]
Multiply \[ - \] inside the bracket in both numerator and denominator,
\[p = \dfrac{{2\sin \theta \left( {1 - \cos \theta - \sin \theta } \right)}}{{1 - 1 - 2\cos \theta \sin \theta }}\]
In denominator \[ + 1\] and \[ - 1\] will get cancel,
\[p = \dfrac{{2\sin \theta \left( {1 - \cos \theta - \sin \theta } \right)}}{{ - 2\cos \theta \sin \theta }}\]
\[2\sin \theta \] is common in both the numerator and denominator, which will get cancel,
\[p = \dfrac{{1 - \cos \theta - \sin \theta }}{{ - \cos \theta }}\]

In denominator there is \[ - \], let multiply both the numerator and denominator by \[ - \],
\[p = \dfrac{{ - (1 - \cos \theta - \sin \theta )}}{{ - ( - \cos \theta )}}\]
\[ - \] in the denominator will get cancel and multiply \[ - \] inside the bracket in numerator,
\[p= \dfrac{{ - 1 + \cos \theta + \sin \theta }}{{\cos \theta }}\]
Rearranging the order in numerators,
\[p = \dfrac{{\cos \theta + \sin \theta - 1}}{{\cos \theta }}\] ………………………………………. (\[1\])
Now simplify \[q\],
\[q = \dfrac{{\cos \theta }}{{1 + \sin \theta }}\]
By rationalizing,
\[q = \dfrac{{\cos \theta }}{{1 + \sin \theta }} \times \dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}\]
\[\Rightarrow q = \dfrac{{\cos \theta (1 - \sin \theta )}}{{(1 + \sin \theta )(1 - \sin \theta )}}\]

Applying \[(a + b)(a - b) = {a^2} - {b^2}\] in denominator,
\[q = \dfrac{{\cos \theta (1 - \sin \theta )}}{{{1^2} - {{\sin }^2}\theta }}\]
We know that, \[{\sin ^2}\theta + {\cos ^2}\theta = 1\], thus \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \], by substituting this,
\[q = \dfrac{{\cos \theta (1 - \sin \theta )}}{{{{\cos }^2}\theta }}\],
\[\cos \theta \], in both numerator and denominator will get cancel,
\[q= \dfrac{{1 - \sin \theta }}{{\cos \theta }}\]
\[\Rightarrow q = \dfrac{{1 - \sin \theta }}{{\cos \theta }}\] …………….…………………………………….. (\[2\])

Adding equation \[1\] and \[2\],
\[p + q = \dfrac{{\cos \theta + \sin \theta - 1}}{{\cos \theta }} + \dfrac{{1 - \sin \theta }}{{\cos \theta }}\]
Here the denominators are same thus,
\[p + q = \dfrac{{\cos \theta + \sin \theta - 1 + 1 - \sin \theta }}{{\cos \theta }}\]
\[ + \sin \theta \] and \[ - \sin \theta \] will get cancel similarly \[ - 1\] and \[ + 1\] will also get cancel,
\[p + q = \dfrac{{\cos \theta }}{{\cos \theta }}\]
\[\cos \theta \] in both numerators and denominators will get cancel and the result will be,
\[\therefore p + q = 1\]

Hence option D is correct.

Note:Rationalization is a method used to simplify the fractions in which the conjugate of the denominator is multiplied with both the numerator and denominator.Conjugate means changing the sign in the middle of two numbers. While simplifying we can find the terms in certain form which already have a predefined formulas like,
\[{(a + b)^2} = {a^2} + 2ab + {b^2}\]
\[\Rightarrow {(a - b)^2} = {a^2} - 2ab + {b^2}\]
\[\Rightarrow (a + b)(a - b) = {a^2} - {b^2}\]
The numbers with the opposite sign will cancel.