We can solve this problem using the basic formula of trigonometric function. Basic formula sin2A+cos2A = 1 cos2A = 1 - sin2A By using this formula we can find the value of k. We have
cosA+cosB
sinA+sinB
= k(
sinA−sinB
cosA−cosB
) (Since‌
a
b
=k
c
d
) ⇒
cosA+cosB
sinA+sinB
×
cosA−cosB
sinA−sinB
= k [Since If
a
b
= k
c
d
,
ad
bc
= k) ⇒
cos2A−cos2B
sin2A−sin2B
= k (Since (a + b) (a - b) = a2−b2) Cross multiplying we get cos2A−cos2B = k(sin2A−sin2B) (1−sin2A)−(1−sin2B) = k(sin2A−sin2B) 1−sin2A−1+sin2B = k(sin2A−sin2B) −sin2A+sin2B = k(sin2A−sin2B) −1(sin2A−sin2B) = k(sin2A−sin2B) i.e., k(sin2A−sin2B) = −1(sin2A−sin2B) ∴ k = - 1 Hence, If