Question

If $p=\left[\frac{2\sin \theta }{(1+\cos \theta +\sin \theta )}\right]$and $q=\left[\frac{\cos \theta }{(1+\sin \theta )}\right]$, then

• A. $pq=1$
• B. $\frac{q}{p}=1$
• C. $q–p=1$
• D. $q+p=1$

Answer 1

The correct option is D

$q+p=1$

Explanation for the correct option.

Step 1: Solve for $p$.

Given,$p=\left[\frac{2\sin \theta }{(1+\cos \theta +\sin \theta )}\right]$

Multiply in numerator and denominator by $1–(cos\theta +\sin \theta )$

$$\begin{gathered} \therefore p=\left[\frac{2\sin \theta }{(1+\cos \theta +\sin \theta )}\right]\times \frac{\left[1–(cos\theta +\sin \theta )\right]}{\left[1–(cos\theta +\sin \theta )\right]} \\ =\frac{2\sin \theta \left[1–(\cos \theta +\sin \theta )\right]}{[1–{(\cos \theta +\sin \theta )}^2]} \\ =\frac{2\sin \theta \left[1–(\cos \theta +\sin \theta )\right]}{[1–({\cos }^2\theta +{\sin }^2\theta +2\cos \theta \sin \theta \left)\right]} \\ =\frac{\left[1–(\cos \theta +\sin \theta )\right]}{-\cos \theta }[{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\mathrm{and}\mathrm{cancelling}\mathrm{out}\mathrm{common}\mathrm{factor}] \\ =\frac{\mathrm{cos\theta }+\mathrm{sin\theta }-1}{\mathrm{cos\theta }}.....\left(1\right) \end{gathered}$$

Step 2: Solve for $q$.

Now solve $q=\left[\frac{\cos \theta }{(1+\sin \theta )}\right]$

Multiply numerator & denominator by $\left(1-\sin \theta \right)$ we have:

$$\begin{gathered} \therefore q=\left[\frac{\cos \theta }{(1+\sin \theta )}\right]\times \frac{(1-\sin \theta )}{(1-\sin \theta )} \\ =\frac{\cos \theta (1-\sin \theta )}{(1-{\sin }^2\theta )}[\left(a+b\right)\left(a-b\right)=a^2-b^2] \\ =\frac{\cos \theta (1-\sin \theta )}{{\cos }^2\theta }[1-{\sin }^2\mathrm{\theta }={\cos }^2\mathrm{\theta }] \\ =\frac{(1-\mathrm{sin\theta })}{\mathrm{cos\theta }}....\left(2\right) \end{gathered}$$

Step 3: Find $p+q$

Adding equation (1) & (2) we get;

$$\begin{gathered} \therefore p+q=\frac{\cos \theta +\sin \theta -1}{\cos \theta }+\frac{1-\sin \theta }{\cos \theta } \\ =\frac{\cos \theta +\sin \theta -1+1-\sin \theta }{\cos \theta } \\ =1 \end{gathered}$$

Therefore, $q+p=1$.

Hence, option D is correct.