If Tan (A + B) = Sqrt3 and Tan (A – B) = 1/Sqrt3 ; 0° < a + B ≤ 90° ; a > B, Find a and B.. - Mathematics

If tan (A + B) = sqrt3 and tan (A – B) = 1/sqrt3 ; 0° < A + B ≤ 90° ; A > B, find A and B.

Solution

tan (A + B) = √3 = tan 60° & tan (A – B) = 1/√3 = tan 30°

A + B = 60° …….(1)

A – B = 30° …….(2)

2A = 90° ⇒ A = 45°

A + B = 60

A – B = 30

Subtract equation (2) from (1)

A + B = 60

A – B = 30

2B = 30°

⇒ B = 15°. Ans.

Note: sin(A + B) = sin A cos B + cos A sin B

sin(A + B) ≠ sin A + sin B.

Concept: Trigonometric Ratios of Some Special Angles
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APPEARS IN

NCERT Class 10 Maths
Chapter 8 Introduction to Trigonometry
Exercise 8.2 | Q 3 | Page 187
RD Sharma Class 10 Maths
Chapter 10 Trigonometric Ratios
Exercise 10.2 | Q 34 | Page 43