Question

show that sin 40 ^0cos70^0=√3.cos80^0​

Answer 1

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L.H.S. :- sin 40 - cos 70                using sin C - sin D 2cos2C+Dsin2C−D  

⇒sin40−cos(90−20)

⇒sin40−sin20 

⇒2cos30.sin10.           sin(90−80)=cos80

⇒223×cos80

=3cos80∘ R.H.S