3sinx-sin3x=4sinx/1+cotx*cotx
Answers
Answered by
2
here is your answer
L.H.S. =
=3sinx - sin3x. (from the formula - Sin3A = 3sinA - 4 sin^3A)
( 3sinA - sin3A = 4sin^3A)
=4sin^3A
=4 sin^2A×sin A.
= 4. ( 1/ cosec^2A) × sinA. ( sin is a reciprocol of cosec )
( so sin A = 1/cosec A)
=4 (1/ 1 + cot ^2A ) × sinA. ( identity cosec ^2 A = 1 + cot^2A)
=4sinA / 1 + cot A *cot A)
= R.H.S.
L.H.S.= R.H.S.
Hence proved
I hope its help you
L.H.S. =
=3sinx - sin3x. (from the formula - Sin3A = 3sinA - 4 sin^3A)
( 3sinA - sin3A = 4sin^3A)
=4sin^3A
=4 sin^2A×sin A.
= 4. ( 1/ cosec^2A) × sinA. ( sin is a reciprocol of cosec )
( so sin A = 1/cosec A)
=4 (1/ 1 + cot ^2A ) × sinA. ( identity cosec ^2 A = 1 + cot^2A)
=4sinA / 1 + cot A *cot A)
= R.H.S.
L.H.S.= R.H.S.
Hence proved
I hope its help you
Similar questions