Question

If α, β are complementary angles such that b sin α = a, then find the value of (sin α cosβ - cos α sinβ).

Answer 1

α, β are complementary angles.
so, the sum of α and β = 90°
e.g., $\alpha+\beta=90^{\circ}$

given, $bsin\alpha=a$
so, $sin\alpha=\frac{a}{b}$
then, $cos\alpha=\frac{\sqrt{b^2-a^2}}{b}$

then, $sin\beta=sin(90-\alpha)=cos\alpha$........(1)
and also, $cos\beta=cos(90-\alpha)=sin\alpha$......(2)

now, $(sin\alpha cos\beta-cos\alpha sin\beta)$

from equations (1) and (2),

= $(sin\alpha sin\alpha - cos\alpha cos\alpha)$

= $sin^2\alpha-cos^2\alpha$

= (a/b)² - {(√b² - a²)/b}²

= a²/b² - (b² - a²)/b²

= (a² - b² + a²)/b²

= (2a² - b²)/b²

Answer 2

HELLO DEAR,


GIVEN:- α and β are the complimentary angle's so, α + β = 90

AND, bsinα = a

=> sinα = a/b

thus, cosα = √{1 - a²/b²}

=> cosα = √{(b² - a²)/b²}

also, sinβ = sin(90 - α) = cosα
similarly, cosβ = sinα

so, sin²α - cos²α

=> (a/b)² - {√(b² - a²)/b}² 

=> a²/b² - (b² - a²)/b² 

=> (a² - b² + a²)/b² 

=> (2a² - b²)/b²


I HOPE IT'S HELP YOU DEAR,
THANKS