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If log tanx=y then prove that sinh(n+1)y+sinh(n−1)y=2sinh ny.cosec2x
1 Answer
written 3.9 years ago by |
**To prove if lntanx=y
sinh(n+1)y+sinh(n−1)y=2sinhnycosec2x**
Now, as lntanx=y we can write,
ey=tanx,
Also by defination of hyperbolic functions
sinhx=ex−e−x2
Hence,
L.H.S. = sinh(n+1)y+sinh(n−1)y
=e(n+1)y−e−(n+1)y2+e(n−1)y−e−(n−1)y2=enyey−e−nye−y2+enye−y−e−nyey2=eny(ey+e−y2)−e−ny(ey+e−y2)=(ey+e−y)(eny−e−ny2)=(tanx+cotx)sinhny
=(sinxcosx+cosxsinx)sinhny=(sin2x+cos2xsinxcosx)sinhny=(1sin2x2)sinhny=(2sin2x)sinhny=2sinhnycosec2x
=R.H.S
Hence Proved.