Question

In the figure above, find tan $$p-$$cot R.

(A) 5
(B) $${\displaystyle \frac{1}{13}}$$
(C) 0
(D) none of these

Answer 1

Hint: To solve this question, we must have an idea about Pythagoras theorem for right-angled triangles.
Here the above picture is a right-angled triangle.
The sides of right-angled triangle named as base, perpendicular, and hypotenuse $$\left[longestside\right]$$ The angle of opposite side of hypotenuse is $${90}^{\circ }.$$

According to Pythagoras theorem,
$${\left(hypotenuse\right)}^2={\left(Perpendicular\right)}^2+{\left(Base\right)}^2$$
$$\Rightarrow$$hypotenuse $$=\sqrt{{\left(perpendicular\right)}^2+{\left(Base\right)}^2}$$
$$\Rightarrow$$ we know that it $$\theta$$ is the angle of a right-angled triangle,
Then, $$\tan \theta ={\displaystyle \frac{perpendicular}{base}}$$
$$\cot \theta ={\displaystyle \frac{1}{\tan \theta }}={\displaystyle \frac{base}{perpendicular}}$$

Complete step by step answer:

Given that
$$PQ=12cm$$
$$PR=13cm$$
$$\mathrm{\angle }PQR={90}^{\circ }$$

At first, we have to find out $$QR$$, the base of $$\mathrm{\Delta }PQR$$
Here, $$PQ=$$ perpendicular to the triangle.
$$QR=$$ base of the triangle.
$$PR=$$ hypotenuse of the triangle.

By Pythagoras theorem,
$$\Rightarrow {\left(hypotenuse\right)}^2={\left(perpendicular\right)}^2+{\left(Base\right)}^2$$
$$\Rightarrow \left(P{R}^2\right)={\left(PQ\right)}^2+{\left(QR\right)}^2$$
$$\Rightarrow {\left(QR\right)}^2={\left(PR\right)}^2-{\left(PQ\right)}^2$$
$$\Rightarrow QR=\sqrt{{\left(PR\right)}^2-{\left(PQ\right)}^2}$$
$$=\sqrt{{\left(13\right)}^2-{\left(12\right)}^2}$$ $$[since,PR=13cm$$
$$\Rightarrow PQ=12cm]$$
$$=\sqrt{169-144}$$
$$=\sqrt{25}=5$$
$$\therefore$$ base $$QR=5cm.$$

Now, we calculate the value of tan p & cot R.
$$\Rightarrow \tan P={\displaystyle \frac{perpendicular}{Base}}$$
$$={\displaystyle \frac{QR}{PQ}}={\displaystyle \frac{5}{12}}$$
 [for angle P, the opposite side of angle is perpendicular $$\left(QR\right).$$Base is $$PQ$$. Opposite side of right angle is hypotenuse]
$$\cot R={\displaystyle \frac{Base}{perpendicular}}$$
$$={\displaystyle \frac{QR}{PQ}}={\displaystyle \frac{5}{12}}$$ [for here angle R, the of angle opposite sides is perpendicular $$\left(PR\right)$$ base is $$QR]$$
$$\therefore \tan P-\cot R={\displaystyle \frac{5}{12}}-{\displaystyle \frac{5}{12}}=0$$

$$\therefore$$ option (c) is right.

Note: A right-angled triangle’s base is one of the sides that adjoins the 90-degree angle.
- The three main functions in trigonometry are sine, cosine, and tangent. It $$\theta$$ is the angle of right-angled triangle then
$$\Rightarrow \sin \theta ={\displaystyle \frac{perpendicular}{hypotenuse}},\cos \theta ={\displaystyle \frac{base}{hypotenuse}}$$ when we solve this type of question, we need to remember all the formulas of trigonometry.