If - and - be the solutions of a cos-b sin-c- then-

The correct options are B sinαsinβ=c^2 a^2/a^2+b^2 C sinα+sinβ=2bc/a^2+b^2 a cosθ=c b sin #10;θ #160; #10;square both sides ⇒ a^2(1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Distance Formula

If α and ...

Question

If $α$ and $β$ be the solutions of $a cos θ + b sin θ = c$ , then-

sin α + sin β = \frac{2 b c}{a^{2} + b^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

sin α sin β = \frac{c^{2} - a^{2}}{a^{2} + b^{2}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

sin α + sin β = \frac{2 a c}{b^{2} + c^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

sin α . sin β = \frac{a^{2} - b^{2}}{b^{2} + c^{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

a x^{2} + b x + c = 0,

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

s i n α + s i n β = a

c o s α + c o s β = b,

s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}

c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}

\pm \sqrt{a^{2} + b^{2} + c^{2}}

\pm \sqrt{a^{2} + b^{2} - c^{2}}

\pm \sqrt{c^{2} - a^{2} - b^{2}}

Join BYJU'S Learning Program