# Trigonometric Ratios CBSE NCERT Solutions Chapter 8 Exercise 8.2 Question 3

Trigonometric Ratios CBSE NCERT Solutions Chapter 8 Exercise 8.2 Question 3

3.   If tan (A+B) = $\sqrt{3}$ and tan(A-B) = $\frac{1}{\sqrt{3}}$; O$^\circ$ (A+B)$\le 90^\circ$, find A and B.

You can check here values of six trigonometric ratios (sin, cos, tan, sec, cot and cosec) for 0, 30, 45, 60 and 90 degrees. You can also learn from this article about how to memorize all the values in an easy way.

Solution:

It is given that tan(A+B) = $\sqrt{3}$  because tan 60$^\circ = \sqrt{3}$

$\Rightarrow (A+B) = 60^\circ$        (1)

It is also given that tan(A-B) = $\frac{1}{\sqrt{3}}$

$\Rightarrow (A-B) = 30^\circ$     because tan 30$^\circ = \frac{1}{\sqrt{3}}$    (2)

Adding (1) and (2), we get

2A = 90

$\Rightarrow A = \frac{90}{2}=45^\circ$

Putting value of A in (1), we get

45 + B = 60

$\Rightarrow B = 60-45 = 15^\circ$

Therefore, A = 45$^\circ$ and B = 15$^\circ$

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