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Question:

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

(i) $(\sqrt{x+1}) \cot \phi$

(ii) $\left(\sqrt{x^{3}+x^{2}}\right) \tan \theta$

(iii) $\cos \theta$

 

Solution:

In $\triangle \mathrm{PQR}, \angle \mathrm{Q}=90^{\circ}$

Using Pythagoras theorem, we get

$\mathrm{PQ}=\sqrt{\mathrm{PR}^{2}-\mathrm{QR}^{2}}$

$=\sqrt{(x+2)^{2}-x^{2}}$

$=\sqrt{x^{2}+4 x+4-x^{2}}$

$=\sqrt{4(x+1)}$

$=2 \sqrt{x+1}$

Now,

(i) $(\sqrt{x+1}) \cot \phi$

$=(\sqrt{x+1}) \times \frac{\mathrm{QR}}{\mathrm{PQ}}$

$=(\sqrt{x+1}) \times \frac{x}{2 \sqrt{x+1}}$

$=\frac{x}{2}$

(ii) $\left(\sqrt{x^{3}+x^{2}}\right) \tan \theta$

$=\left(\sqrt{x^{2}(x+1)}\right) \times \frac{\mathrm{QR}}{\mathrm{PQ}}$

$=x \sqrt{(x+1)} \times \frac{x}{2 \sqrt{x+1}}$

$=\frac{x^{2}}{2}$

(iii) $\cos \theta$

$=\frac{\mathrm{PQ}}{\mathrm{PR}}$

$=\frac{2 \sqrt{x+1}}{(x+2)}$

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