Question

Prove : sin theta/1-cos theta = cosec
theta + cot theta

Answer 1

Question :

Prove :

$\bf \dfrac{ \sin \theta}{1 - \cos \theta} = \cosec \theta + \cot \theta$

Trignometric Formula's:

sin²A + cos²A = 1

sec²A - tan²A = 1

cosec²A - cot²A = 1

Solution :

$\bf \dfrac{ \sin \theta}{1 - \cos \theta} = \cosec \theta + \cot \theta$

LHS

$= \sf \dfrac{ \sin \theta}{1 - \cos \theta}$

$= \sf \dfrac{ \sin \theta}{1 - \cos \theta} \times \dfrac{1 + \cos \theta}{1 - \cos \theta}$

$= \dfrac{ \sin \theta(1 + \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta) }$

We know that $\bf\:(a+b)(a-b)=a{}^{2}-b{}^{2}$

$= \dfrac{ \sin \theta(1 + \cos \theta)}{1 - \cos {}^{2} \theta }$

Use formula sin²A + cos²A = 1

$= \dfrac{ \sin \theta(1 + \cos \theta)}{ \sin { }^{2} \theta}$

$= \dfrac{ \cancel{\sin \theta}(1 + \cos \theta)}{ \cancel{ \sin \theta} \times \sin \theta}$

$= \sf \dfrac{(1 + \cos \theta)}{ \sin \theta}$

$= \sf \cosec \theta + \cot \theta$

RHS

$= \sf \cosec \theta + \cot \theta$

⇒LHS = RHS

$\huge{\bold{ Hence\:Proved}}$

_____________________________

More Trigonometric Formula's:

sin2A = 2 sinA cosA

cos2A = cos²A - sin²A

tan2A = 2 tanA / (1 - tan²A)

Answer 2

${\boxed{\mathsf{\purple{Solution}}}}$

${\boxed{\mathsf{To\:Prove}}}$

$\implies \frac{ \sin \theta}{1 - \cos \theta} = \cosec \theta + \cot \theta$

Taking LHS →

$\implies \frac{ \sin \theta}{1 - \cos \theta}$

Now multiplying and dividing the fraction by 1 + cos theta .

$\implies \frac{ \sin \theta}{1 - \cos \theta} \times \frac{1 + \cos \theta}{1 + \cos \theta}$

$\frac{ 1 }{ \sin \theta} + \frac{ \cos \theta}{ \sin \theta}$

As we know that ( a + b ) × ( a - b ) = a² - b² . So ,

$\implies \frac{ \sin \theta + \sin \theta \times \cos \theta}{{1}^{2} - {\cos \theta}^{2}}$

Now let's have look over some trigonometric identities :-

→ Sin²A + Cos²A = 1

→ 1 - Cos²A = Sin²A .

Now replacing 1 - Cos²A by Sin²A .

$\implies \frac{ \sin \theta + \sin \theta . \cos \theta}{{\sin \theta}^{2}}$

$\implies \frac{ \sin \theta}{ {\sin \theta}^{2} } + \frac{\sin \theta . \cos \theta}{ {\sin \theta}^{2} }$

$\implies \frac{1}{ \sin \theta } + \frac{ \cos \theta}{ \sin \theta}$

As we know that 1/SinA = Cosec A and

CosA/SinA = Cot A

$\implies \cosec \theta + \cot \theta$

Hence proved .