A sin x-bsinx-2-3- x sinx-4-3 find the ab-bc-ca-

Let, asin x=bsin( x+ 2π #160; 3 #160; ) =csin( x+ 4π #160; 3 #160; ) =k hence, k a =sin x; k b =sin( x+ 2π #160; 3 #160 ...

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Byju's Answer

Standard XII

Mathematics

Relation between Inverses of Trigonometric Functions and Their Reciprocal Functions

a sin x=bsinx...

Question

$a sin x = b sin (x + \frac{2 π}{3}) = x sin (x + \frac{4 π}{3})$ find the $a b + b c + c a = ?$

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Solution

$L e t, a sin x = b sin (x + \frac{2 π}{3}) = c sin (x + \frac{4 π}{3}) = k h e n c e, \frac{k}{a} = sin x; \frac{k}{b} = sin (x + \frac{2 π}{3}); \frac{k}{c} = sin (x + \frac{4 π}{3}) \Rightarrow \frac{k}{c} = sin (2 π + (x - \frac{2 π}{3})) = sin (x - \frac{2 π}{3}) ∴ \frac{k}{a} + \frac{k}{b} + \frac{k}{c} = sin x + sin (x + \frac{2 π}{3}) + sin (x - \frac{2 π}{3}) \Rightarrow k (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = sin x + sin x cos \frac{2 π}{3} + sin \frac{2 π}{3} cos x + sin x cos \frac{2 π}{3} - sin \frac{2 π}{3} cos x \Rightarrow k . \frac{a b + b c + c a}{a b c} = sin x + 2 cos \frac{2 π}{3} sin x \Rightarrow k . \frac{a b + b c + c a}{a b c} = 0 (∵ cos \frac{2 π}{3} = - \frac{1}{2}) s i n c e k i s n o t z e r o ∴ \frac{a b + b c + c a}{a b c} = 0 \Rightarrow a b + b c + c a = 0$

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asinx=bsin(x+2π3)=xsin(x+4π3) find the ab+bc+ca=?

$a sin x = b sin (x + \frac{2 π}{3}) = x sin (x + \frac{4 π}{3})$ find the $a b + b c + c a = ?$