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

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## Question

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

Video Solution
Introduction To Trigonometry
Ex 8.1 | Question 2

## Text Solution

#### What is the known?

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

#### What is the unknown?

One side of right-angled triangle $$\Delta PQR$$

#### Reasoning:

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

#### Steps:

Apply Pythagoras theorem for $$\Delta PQR$$ we obtain:

\begin{align} 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 \end{align}

\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}}\\ &= \frac{\text{QR}}{\text{PQ}} \\ & = \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}} \end{align}

\begin{align} \rm {\tan} \,{P}- {{cot}}\,{R} & = \frac{{5}}{{12}}-\frac{{5}}{{12}} \\ \rm{tan}\,P- {cot}\,{R} &= {0}\end{align}

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