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Q.

State true or false.
$\frac{1 - cosx}{sinx} = \tan \frac{x}{2}$ ?

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a

True

b

False

answer is A.

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Detailed Solution

\frac{1 - cosx}{sinx} = \tan \frac{x}{2}

LHS

= \frac{1 - cosx}{sinx}

= \frac{2 si n^{2} \frac{x}{2}}{sin x} [1 - cosθ = 2 si n^{2} \frac{θ}{2}]

= \frac{2 si n^{2} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}

[sin θ = 2 \sin \frac{θ}{2} \cos \frac{θ}{2}]

= \frac{2 si n^{} \frac{x}{2} si n^{} \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}

= \frac{si n^{} \frac{x}{2}}{\cos \frac{x}{2}}

= \tan \frac{x}{2}

. ……..1 [

\tan θ = \frac{sin θ}{cos θ}

]
RHS

= \tan \frac{x}{2}

……….2
From, 1 & 2
LHS=RHS.

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