Class X Session 2023-24 Question 24

 MATH SAMPLE QUESTION PAPER

Class X Session 2023-24

MATHEMATICS STANDARD (Code No.041) 

Question 24

$$if\tan (A+B)=\sqrt{3}and\tan (A-B)=\frac{1}{\sqrt{3}}$$ ,0° < A + B < 90°; A > B, find A and B.

Explanation:  

Given that :

$$\tan (A+B)=\sqrt{3}and\tan (A-B)=\frac{1}{\sqrt{3}}$$

First, recognize the angles for which

$$\tan (\theta )=\sqrt{3}andtan(\theta )=\frac{1}{\sqrt{3}}$$ we know that

$$\tan ({60}^{\circ })=\sqrt{3}andtan({30}^{\circ })=\frac{1}{\sqrt{3}}$$

$$so,A+B={60}^{\circ }andA-B={30}^{\circ }$$ now add the two equation , we get 

$$A+B={60}^{\circ },---(1)$$ 

$$A-B={30}^{\circ },---(2)$$

$$so,(A+B)+(A-B)={60}^{\circ }+{30}^{\circ }$$

$$or,2A={90}^{\circ }$$

$$or,A={45}^{\circ }$$

Now Put A value in equation (1) and we get 

$$from(1),B={60}^{\circ }-{45}^{\circ }$$

$$\therefore B={15}^{\circ }$$

$$NowA={45}^{\circ }andB={15}^{\circ }$$