Question

Prove that
sin 0/(1 - cos 0)+cos0/ (1 - tan )
= cos 0+ sin 0​

Answer 1

Step-by-step explanation:

this is the proof for your question

Answer 2

Appropriate Question

$\bf\bullet\ \frac{sin\theta}{1-cot\theta} + \frac{cos\theta}{1-tan\theta} = cos\theta + sin\theta$

Step-by-step explanation:

According to the Question

$\sf\bullet\ \frac{sin\theta}{1-cot\theta} + \frac{cos\theta}{1-tan\theta} = cos\theta + sin\theta$

Taking LHS

$\dashrightarrow\bf \frac{sin^{2} \theta}{sin(1-cot\theta)} + \frac{cos^{2}\theta}{cos(1-tan\theta)}$

Now , we know that

• tanθ  = sinθ/cosθ

• cotθ = cosθ/sinθ

by substituting this value we get

$\dashrightarrow\bf\; \frac{sin^2\theta}{sin\theta-cos\theta} + \frac{cos^2\theta}{cos\theta-sin\theta} \\\\\\\dashrightarrow\bf\; \frac{sin^2\theta}{sin\theta-cos\theta} - \frac{cos^2\theta}{sin\theta-cos\theta} \\\\\\\dashrightarrow\bf\; \frac{sin^2\theta-cos^2\theta}{sin\theta-cos\theta}$

Now , using identity to split the numerator .

•(a²-b²) = (a+b) (a-b)

$\dashrightarrow\bf\; \frac{(sin\theta+cos\theta)(sin\theta-cos\theta)}{(sin\theta-cos\theta)} \\\\\dashrightarrow\bf\; (sin\theta+cos\theta) \\\\\dashrightarrow\bf\; sin\theta+cos\theta$

LHS = RHS

Hence , Proved !!