Question
If $\sin\theta+\cos\theta=\text{p and }\sec\theta+\text{cosec}\theta=\text{q},$ then prove that $\text{q}(\text{p}^2-1)=2\text{p}.$
✓
Answer
$\sin\theta+\cos\theta=\text{p}\ \ ...(\text{I})$
$\sec\theta+\text{cosec}\theta=\text{q}\ \ ...(\text{II})$
[(II)nd expretion can be changed into $\sin\theta,\ \cos\theta$ and eliminate trigonometric ratio from (I) and (II)]
$\sec\theta+\text{cosec}\theta=\text{q}$
$\Rightarrow\ \frac{1}{\cos\theta}+\frac{1}{\sin\theta}=\text{q}$
$\Rightarrow\ \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac{\text{q}}{1}$
$\Rightarrow\ \frac{\text{p}}{\sin\theta\cos\theta}=\text{q}\ \ [\text{Using(I)}]$
$\Rightarrow\ \sin\theta\cos\theta=\frac{\text{p}}{\text{q}}\ \ (\text{III})$
$\sin\theta+\cos\theta=\text{p}\ \ [\text{From(I)}]$
$\Rightarrow(\sin\theta+\cos\theta)^2=\text{p}^2\ \ [\text{squaring both sides}]$
$\Rightarrow\ \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=\text{p}^2$
$\Rightarrow\ 1+2\cdot\frac{\text{p}}{\text{q}}=\text{p}^2\ \ [\because\sin^2\theta+\cos^2\theta=1\text{ and using (III)}]$
$\Rightarrow\ \text{q}+2\text{p}=\text{p}^2\text{q}$
$\Rightarrow\ 2\text{p}=\text{p}^2\text{q}-\text{q}$
$\Rightarrow\ 2\text{p}=\text{q}(\text{p}^2-1)$
Hence, proved.
$\sec\theta+\text{cosec}\theta=\text{q}\ \ ...(\text{II})$
[(II)nd expretion can be changed into $\sin\theta,\ \cos\theta$ and eliminate trigonometric ratio from (I) and (II)]
$\sec\theta+\text{cosec}\theta=\text{q}$
$\Rightarrow\ \frac{1}{\cos\theta}+\frac{1}{\sin\theta}=\text{q}$
$\Rightarrow\ \frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\frac{\text{q}}{1}$
$\Rightarrow\ \frac{\text{p}}{\sin\theta\cos\theta}=\text{q}\ \ [\text{Using(I)}]$
$\Rightarrow\ \sin\theta\cos\theta=\frac{\text{p}}{\text{q}}\ \ (\text{III})$
$\sin\theta+\cos\theta=\text{p}\ \ [\text{From(I)}]$
$\Rightarrow(\sin\theta+\cos\theta)^2=\text{p}^2\ \ [\text{squaring both sides}]$
$\Rightarrow\ \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=\text{p}^2$
$\Rightarrow\ 1+2\cdot\frac{\text{p}}{\text{q}}=\text{p}^2\ \ [\because\sin^2\theta+\cos^2\theta=1\text{ and using (III)}]$
$\Rightarrow\ \text{q}+2\text{p}=\text{p}^2\text{q}$
$\Rightarrow\ 2\text{p}=\text{p}^2\text{q}-\text{q}$
$\Rightarrow\ 2\text{p}=\text{q}(\text{p}^2-1)$
Hence, proved.
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