Ex.8.1 Q2 Introduction to Trigonometry Solution - NCERT Maths Class 10

Go back to  'Ex.8.1'


In the given figure, find \(\text{tan}\,\text{P }-\text{cot }\text{R}.\)

Text Solution

What is the known?

\(\rm{PQ} = 12\,\rm{ cm}\) and \(\rm{PR} = 13\,\rm {cm}.\)

What is the unknown?

One side of right-angled triangle \(\Delta \text{PQR}\)


Using Pythagoras theorem, we can find the length of the third side, then the required trignometric ratios.


Apply Pythagoras theorem for \(\Delta \rm{PQR}\) we obtain:

P{R^2}\, &= \,P{Q^2} + Q{R^2}\\
Q{R^2}\, &= \,P{R^2} - P{Q^2}\\
Q{R^2} &= {(13\rm cm)^2}\, - \,{(12\rm cm)^2}\\
Q{R^2} &= 169\,\rm c{m^2} - 144\,\rm c{m^2}\\
Q{R^2} &= 25\,\rm c{m^2}\\
QR\, &= \,5\,\rm cm

\[\begin{align} & \text{tan}\,\text{P = }\frac{\text{side}\ \text{opposite}\ \text{to}\ \angle \text{P}}{\text{side}\ \text{adjacent}\ \text{to}\ \angle \text{P}}\text{= }\frac{\text{QR}}{\text{PQ}}\text{= }\frac{\text{5}\,\text{cm}}{\text{12}\,\text{cm}} \\ & \text{tan}\,\text{P = }\frac{\text{5}}{\text{12}} \end{align}\]

\[\begin{align} \rm{cot}\,{R} &= \frac{ \text{side adjacent to}\ \angle {R}}{\text{side opposite to}\ \angle {R}}{= }\frac{{QR}}{{PQ}}\ \,=\frac{{5}\,\rm{cm}}{{12}\,\rm{cm}} \\ \rm {cot}\,{R} &= \frac{{5}}{{12}} \\\rm {\tan} \,{P}- {{cot}}\,{R} & = \frac{{5}}{{12}}-\frac{{5}}{{12}} \\ \rm{tan}\,P- {cot}\,{R} &= {0}\end{align}\]