In the figure of , = ^ and = ^ . Find: 1. ( x+1 ) 2. ( x^3+x^2 ) … - Vidyadip

Question

In the figure of $\triangle\text{PQR},\angle\text{P}=\theta^\circ$and $\angle\text{R}=\phi^\circ.$
Find:

$\big(\sqrt{\text{x}+1}\big)\cot\phi$
$\big(\sqrt{\text{x}^3+\text{x}^2}\big)\tan\theta$
$\cos\theta$

✓

Answer

In $\triangle\text{PQR},\angle\text{Q}=90^\circ,\angle\text{P}=\theta^\circ$ and $\angle\text{R}=\phi^\circ$
By Pythagoras theorem, we have
$\text{PQ}^2=\text{PR}^2-\text{QR}^2$
$=\big(\text{x}+2\big)^2-\text{x}^2=\text{x}^2+4\text{x}+4-\text{x}^2=4(\text{x}+1)$
$\Rightarrow\text{PQ}=2\sqrt{\text{x}+1}$
Now, $\cot\phi=\frac{\text{QR}}{\text{PQ}}=\frac{\text{x}}{2\sqrt{\text{x}+1}}$ and $\tan\theta=\frac{\text{QR}}{\text{PQ}}=\frac{\text{x}}{2\sqrt{\text{x}+1}}$
$\big(\sqrt{\text{x}+1}\big)\cot\phi=\big(\sqrt{\text{x}+1}\big)\times\frac{\text{x}}{2\sqrt{\text{x}+1}}=\frac{\text{x}}{2}$
$\big(\sqrt{\text{x}^3+\text{x}^2}\big)\tan\theta=\big(\sqrt{\text{x}^2(\text{x}+1)}\big)\\\tan\theta=\text{x}\big(\sqrt{\text{x}+1}\big)\times\frac{\text{x}}{2\sqrt{\text{x}+1}}=\frac{\text{x}^2}{2}$
$\cos\theta=\frac{\text{PQ}}{\text{PR}}=\frac{2\sqrt{\text{x}+1}}{\text{x}+2}$

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