Question

8. If A = B = 45°, show that :
(i) sin (A - B) = sin A cos B - cos A sin B.
​

Answer 1

Given :-

A = B = 45°

To prove :-

sin (A - B ) = sinA cosB - cosA sinB

To know :-

• sin0° = 0
• sin45° = 1/√2
• cos45 = 1/√2

Solution :-

Substitute value of A,B in given equation

A = 45° , B = 45°

sin (A - B ) = sinA cosB - cosA sinB

sin (45-45) = (sin45 × cos45) - (cos45 × sin45)

sin0° =( 1/√2 × 1/√2 ) - ( 1/√2 × 1/√2 )

0 = 1/2 - 1/2

0 = 0

Hence LHS = RHS

Proved !!

Know more:-

cos(A + B) = cosAcosB - sinAsinB

cos ( A - B) = cosA cosB + sinAsinB

tan ( A +B ) = tanA + tanB / 1 - tanAtanB

tan( A-B) = tanA - tanB/1 + tanAtanB

cot ( A + B ) = cotBcotA -1 / cotB + cotA

cot ( A - B ) = cotB cotA + 1/ cotB - cotA

tan(45+ A) = 1+tanA/1 - tanA

tan (45 - A ) = 1-tanA/1 + tanA

Trignometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trignometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trignometric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj