Question

Show that (cosec 0 - cot 0)^2=
1 -- cos 0by
1+ cos 0​

Answer 1

Step-by-step explanation:

Let's represent your question first,

${( cosec \: θ - cot \:θ)}^{2} = \frac{1 - cos \: θ}{1 + cos \:θ}$

Consider the RHS of the equation (that is, what is there to the right side of the 'equal to' sign)

Divide both the numerator and denominator by sinθ, since we are dividing both the numerator and the denominator by the same quantity, it will not affect the original question.

$RHS = \frac{1 - cos \:θ}{1 + \cosθ} = \frac{1 - cos \:θ}{1 + \cosθ} \div 1 \\ = \frac{1 - cos \:θ}{1 + \cosθ} \div \frac{ \sinθ }{ \sinθ } = \frac{ \frac{1}{sin \:θ} - \frac{ \cosθ }{ \sinθ } }{\frac{1}{sin \:θ} + \frac{ \cosθ }{ \sinθ }} \\ = \frac{cosec \:θ - cot \:θ}{cosec \:θ + cot \:θ}$

Now, multiply the obtained fraction with (cosec θ - cot θ) on both the numerator and the denominator.

$\frac{cosec \:θ - cot \:θ}{cosec \:θ + cot \:θ} \times \frac{cosec \:θ - cot \:θ}{cosec \:θ - cot \:θ} = \\ \frac{ {(cosec \:θ - cot \:θ)}^{2} }{{cosec}^{2} \:θ - {cot}^{2} \:θ}$

But we know cosec² θ - cot² θ = 1 (Trigonometric identity)

So the denominator becomes 1.

$\frac{ {(cosec \:θ - cot \:θ)}^{2} }{{cosec}^{2} \:θ - {cot}^{2} \:θ} = \frac{ {(cosec \:θ - cot \:θ)}^{2} }{1} = LHS$

LHS = RHS

Hence, proved.

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