Question

prove that sin(50+Theta) - cos(40- theta) = 0

Answer 1

LHS,



=>Sin(50+∅) - cos(40-∅)

$\cos(theta) = \sin(90 - theta)$


=>Sin(50+∅)- sin{90-(40-∅)}


=>sin(50+∅) - sin{90-40+∅)

=>Sin(50+∅) - sin(50+∅)

=0

RHS,

=> 0

---------------------

0=0
LHS=RHS



I hope this will help you


-by ABHAY

Answer 2

Here your answer goes

Step :- 1

Given ,

Sin 50° can be written as 90° - 45°

⇒ Sin (90 - 40 + theta ) - Cos ( 40 - theta )

Step :- 2

⇒ sin(90-40+theta) - cos(40-theta)

⇒ sin[90-(40-theta)]- cos(40-theta)

⇒ $cos(40-theta)-cos(40-theta)$

∴  sin ( 90 - theta ) = Cos theta

⇒ 0

Therefore ,  Sin ( 50 + theta ) - Cos( 40 - theta ) = 0

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