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Question: If \[p = \left[ {2\sin \theta /\left( {1 + \cos \theta + \sin \theta } \right)} \right]\] and \[q = ...

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

Explanation

Solution

In order to Simplify the values of pp and qq 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⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1.

Complete step by step answer:
Given, p=2sin⁡θ1+cos⁡θ+sin⁡θp = \dfrac{{2\sin \theta }}{{1 + \cos \theta + \sin \theta }} and q=cos⁡θ1+sin⁡θq = \dfrac{{\cos \theta }}{{1 + \sin \theta }}.
Let simplify pp,
p=2sin⁡θ1+cos⁡θ+sin⁡θp = \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=2sin⁡θ1+(cos⁡θ+sin⁡θ)p = \dfrac{{2\sin \theta }}{{1 + (\cos \theta + \sin \theta )}}
Now rationalize,
p=2sin⁡θ1+(cos⁡θ+sin⁡θ)×1−(cos⁡θ+sin⁡θ)1−(cos⁡θ+sin⁡θ)p= \dfrac{{2\sin \theta }}{{1 + (\cos \theta + \sin \theta )}} \times \dfrac{{1 - (\cos \theta + \sin \theta )}}{{1 - (\cos \theta + \sin \theta )}}
⇒p=2sin⁡θ(1−(cos⁡θ+sin⁡θ))(1+(cos⁡θ+sin⁡θ))(1−(cos⁡θ+sin⁡θ))\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)(a + b)(a - b), by applying the formula: (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2},
⇒p=2sin⁡θ(1−(cos⁡θ+sin⁡θ))12−(cos⁡θ+sin⁡θ)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{(a + b)^2}, by applying the formula: (a+b)2=a2+b2+2ab{(a + b)^2} = {a^2} + {b^2} + 2ab,
p=2sin⁡θ(1−(cos⁡θ+sin⁡θ))12−(cos⁡2θ+sin⁡2θ+2cos⁡θsin⁡θ)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 12=1{1^2} = 1 and sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 by substituting the values,
p=2sin⁡θ(1−(cos⁡θ+sin⁡θ))1−(1+2cos⁡θsin⁡θ)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=2sin⁡θ(1−cos⁡θ−sin⁡θ)1−1−2cos⁡θsin⁡θp = \dfrac{{2\sin \theta \left( {1 - \cos \theta - \sin \theta } \right)}}{{1 - 1 - 2\cos \theta \sin \theta }}
In denominator +1 + 1 and −1 - 1 will get cancel,
p=2sin⁡θ(1−cos⁡θ−sin⁡θ)−2cos⁡θsin⁡θp = \dfrac{{2\sin \theta \left( {1 - \cos \theta - \sin \theta } \right)}}{{ - 2\cos \theta \sin \theta }}
2sin⁡θ2\sin \theta is common in both the numerator and denominator, which will get cancel,
p=1−cos⁡θ−sin⁡θ−cos⁡θp = \dfrac{{1 - \cos \theta - \sin \theta }}{{ - \cos \theta }}

In denominator there is −-, let multiply both the numerator and denominator by −-,
p=−(1−cos⁡θ−sin⁡θ)−(−cos⁡θ)p = \dfrac{{ - (1 - \cos \theta - \sin \theta )}}{{ - ( - \cos \theta )}}
−- in the denominator will get cancel and multiply −- inside the bracket in numerator,
p=−1+cos⁡θ+sin⁡θcos⁡θp= \dfrac{{ - 1 + \cos \theta + \sin \theta }}{{\cos \theta }}
Rearranging the order in numerators,
p=cos⁡θ+sin⁡θ−1cos⁡θp = \dfrac{{\cos \theta + \sin \theta - 1}}{{\cos \theta }} ………………………………………. (11)
Now simplify qq,
q=cos⁡θ1+sin⁡θq = \dfrac{{\cos \theta }}{{1 + \sin \theta }}
By rationalizing,
q=cos⁡θ1+sin⁡θ×1−sin⁡θ1−sin⁡θq = \dfrac{{\cos \theta }}{{1 + \sin \theta }} \times \dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}
⇒q=cos⁡θ(1−sin⁡θ)(1+sin⁡θ)(1−sin⁡θ)\Rightarrow q = \dfrac{{\cos \theta (1 - \sin \theta )}}{{(1 + \sin \theta )(1 - \sin \theta )}}

Applying (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2} in denominator,
q=cos⁡θ(1−sin⁡θ)12−sin⁡2θq = \dfrac{{\cos \theta (1 - \sin \theta )}}{{{1^2} - {{\sin }^2}\theta }}
We know that, sin⁡2θ+cos⁡2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1, thus cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta , by substituting this,
q=cos⁡θ(1−sin⁡θ)cos⁡2θq = \dfrac{{\cos \theta (1 - \sin \theta )}}{{{{\cos }^2}\theta }},
cos⁡θ\cos \theta , in both numerator and denominator will get cancel,
q=1−sin⁡θcos⁡θq= \dfrac{{1 - \sin \theta }}{{\cos \theta }}
⇒q=1−sin⁡θcos⁡θ\Rightarrow q = \dfrac{{1 - \sin \theta }}{{\cos \theta }} …………….…………………………………….. (22)

Adding equation 11 and 22,
p+q=cos⁡θ+sin⁡θ−1cos⁡θ+1−sin⁡θcos⁡θp + q = \dfrac{{\cos \theta + \sin \theta - 1}}{{\cos \theta }} + \dfrac{{1 - \sin \theta }}{{\cos \theta }}
Here the denominators are same thus,
p+q=cos⁡θ+sin⁡θ−1+1−sin⁡θcos⁡θp + q = \dfrac{{\cos \theta + \sin \theta - 1 + 1 - \sin \theta }}{{\cos \theta }}
+sin⁡θ+ \sin \theta and −sin⁡θ- \sin \theta will get cancel similarly −1 - 1 and +1 + 1 will also get cancel,
p+q=cos⁡θcos⁡θp + q = \dfrac{{\cos \theta }}{{\cos \theta }}
cos⁡θ\cos \theta in both numerators and denominators will get cancel and the result will be,
∴p+q=1\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=a2+2ab+b2{(a + b)^2} = {a^2} + 2ab + {b^2}
⇒(a−b)2=a2−2ab+b2\Rightarrow {(a - b)^2} = {a^2} - 2ab + {b^2}
⇒(a+b)(a−b)=a2−b2\Rightarrow (a + b)(a - b) = {a^2} - {b^2}
The numbers with the opposite sign will cancel.