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|NCERT| |Class 11| |Trigonometric Function|

Prove that :

$\frac{ \sin5x - 2 \sin3x + \sin \: x}{cos5x - cosx} = \tan \: x$

Thanks !

Answer 1

we know that sinA + sinB = 2sin(A + B)/2 × cos(A - B)/2

⇒ sin5x + sinx = 2sin6x/2 × cos4x/2 = 2sin3x × cos2x

we know that cosA - cosB = -2sin(A + B)/2 × sin(A - B)/2

cos5x - cosx = -2sin3x × sin2x

now substituting the above values in the question we get

⇒ - {2sin3x × cos2x - 2sin3x}/2sin3x × sin2x

taking - 2sin3x common in the numerator we get

⇒{2sin3x(1 - cos2x)}/2sin3x × sin2x

2sin3x gets cancelled in numerator and denominator we get

⇒ {1 - cos2x}/sin2x

but we know that {1 - cos2x} = 2sin²x and sin2x = 2sinxcosx

substituiting we get 2sin²x/2sinxcosx

2sinx gets cancelled in numerator and denominator

⇒ we get sinx/cosx = tanx

Answer 2

$\frac{ \sin5x - 2 \sin3x + \sin \: x}{cos5x - cosx} = \tan \: x$