Question

If a coa teta - b sin teta = c , prove that ( a sin teta + b coa teta ) = +root a² + b² - c² , -root a²+b²-c².

Answer 1

Hey !!


Given : a cos ¢ - b sin ¢ = C


Now,


( a cos ¢ - b sin ¢ )² + ( a sin ¢ + b cos¢)²


=> a² ( cos²¢ + sin²¢ ) + b² ( sin²¢ + cos² ¢ ) = ( a² + b² ).



Thus,


( a cos ¢ - b sin¢)² + ( a sin¢ + b cos¢)² = ( a²+b²).



=> C² + ( a sin ¢ + b cos ¢ )² = ( a² + b²).


=> ( a sin ¢ + b cos ¢ )² = ( a² + b² + c²)



=> ( a sin¢ + b cos ¢ ) = +- √ a²+ b² - c²



Hence,


( a sin ¢ + b cos ¢ ) = +- ✓ a² + b² - c²

Answer 2

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